c assignment of read only location

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The qualifier const can be applied to the declaration of any variable to specify that its value will not be changed (which depends upon where const variables are stored, we may change the value of the const variable by using a pointer). The result is implementation-defined if an attempt is made to change a const.

Using the const qualifier in C is a good practice when we want to ensure that some values should remain constant and should not be accidentally modified.

In C programming, the const qualifier can be used in different contexts to provide various behaviors. Here are some different use cases of the const qualifier in C:

1. Constant Variables

In this case, const is used to declare a variable var as a constant with an initial value of 100. The value of this variable cannot be modified once it is initialized. See the following example:

2. Pointer to Constant

We can change the pointer to point to any other integer variable, but cannot change the value of the object (entity) pointed using pointer ptr. The pointer is stored in the read-write area (stack in the present case). The object pointed may be in the read-only or read-write area. Let us see the following examples.

Example 2: Program where variable i itself is constant.

Down qualification is not allowed in C++ and may cause warnings in C. Down qualification refers to the situation where a qualified type is assigned to a non-qualified type.

Example 3: Program to show down qualification.

3. Constant Pointer to Variable

The above declaration is a constant pointer to an integer variable, which means we can change the value of the object pointed by the pointer, but cannot change the pointer to point to another variable.

4. Constant Pointer to Constant

The above declaration is a constant pointer to a constant variable which means we cannot change the value pointed by the pointer as well as we cannot point the pointer to other variables. Let us see with an example. 

Advantages of const Qualifiers in C

The const qualifier in C has the following advantages:

• Improved Code Readability: By marking a variable as const, you indicate to other programmers that its value should not be changed, making your code easier to understand and maintain.
• Enhanced Type Safety : By using const, you can ensure that values are not accidentally modified, reducing the chance of bugs and errors in your code.
• Improved Optimization: Compilers can optimize const variables more effectively, as they know that their values will not change during program execution. This can result in faster and more efficient code.
• Better Memory Usage: By declaring variables as const, you can often avoid having to make a copy of their values, which can reduce memory usage and improve performance.
• Improved Compatibility : By declaring variables as const, you can make your code more compatible with other libraries and APIs that use const variables.
• Improved Reliability : By using const, you can make your code more reliable, as you can ensure that values are not modified unexpectedly, reducing the risk of bugs and errors in your code.

This article is compiled by “ Narendra Kangralkar “.

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Const Pointer in C++

In this lecture, we discuss about Const pointer and Pointer to Const and Const Pointer to Const in C++.

Why do we need Const Pointer or Pointer to Const?

A pointer is a variable that stores the memory address of another variable or the memory address of the memory allocated on the heap. For example,

Here, we have created a variable ‘num’ and initialized it with the value 11, then created a pointer ‘ptr’ and initialized it with the address of variable ‘num’. Now, the pointer ‘ptr’ contains the memory location where the value of the variable ‘num’, i.e. 11, is stored.

You can access this value using the pointer ‘ptr’ by dereferencing it. Like this,

You can also change the value at that memory location. For example, you could change the value at the memory location pointed to by ‘ptr’ to 20, which in turn would change the value of variable ‘num’. You can print and confirm that the value of the variable ‘num’ is now 20.

This is how you change the value at a memory location using a pointer.

Now, suppose you pass a pointer to a function and, inside the function, someone changes the value. This value change inside the function using the pointer will be reflected outside the function as well. For example, if you have a variable ‘num’, initialized with the value 11, and then create a pointer ‘ptr’ containing the address of this variable, if you then pass this pointer to the function ‘getSquare’, this function is supposed to return the square of the value at the memory location pointed to by the given pointer. But if a new developer comes along and tries to change the value at this location pointed to by the pointer, because you passed a pointer, he can directly change the value.

We wanted the square of 11, but inside the function value at memory location got changed to 2, and it returned the square of 2 instead. Also after the function getSquare() returns, the value of num is also changed to 2.

Pointer to Const in C++

You can restrict the user from changing the value pointed to by the pointer by using a pointer to const. While creating a pointer, you can make the pointer point to a const integer. In both cases, the pointer ‘ptr’ will be pointing to a memory location, but it cannot change the data at that memory location. So, if someone tries to change the value now, it will give an error. Similarly, if you want to pass a pointer to a function for read operation only, then you can pass it like this.

Here, you are passing a pointer pointing to a const integer, so nobody can change the value at that memory location using the pointer. This is called pointer to const.

Even if you restricted the user so that they cannot change the value at the memory location pointed to by the pointer ‘ptr’, the pointer itself is also a variable, so you can change it to point to a new memory location. Check out this example: here, inside the ‘getSquare()’ function, we change the value of the pointer ‘ptr’ and make it point to a new memory location.

Is there any way to restrict the user so that they can’t change the pointer to point to a new memory location? Answer is Yes, by making pointer a constant pointer.

To restrict the user so they cannot make the pointer point to another memory location, you need to make the pointer const, like this.

Here, your pointer is a const pointer pointing to a memory location, or the address of variable ‘num’. Now, even if you want to point the pointer to a new memory location, you can’t do that because the pointer is of const type. However, using the pointer, you can change the value at that memory location, as the pointer is not pointing to a constant integer. This is different from the previous scenario.

Const pointer pointing to Const in C++

In this section, we will cover a combination of both: a const pointer pointing to a constant integer. Suppose you initialize the variable ‘num’ with the value 11. Now, you want to create a pointer to this but you want to restrict it so that no one can change the value of ‘num’ using the pointer, and also, once you initialize the pointer to the address of ‘num’, no one can make that pointer point to any other location. For that, you have to use a const pointer pointing to a const value, like this.

Here, your pointer ‘ptr’ is a const pointer and is pointing to a constant value. So, we cannot make the pointer point to any other location, and we also cannot use the pointer to change the value at that memory location.

Const pointer vs Pointer to Const vs Const Pointer to Const

Sure, here’s a simple table to illustrate the differences:

• A const pointer means that the memory location that the pointer is pointing to can be changed, but the pointer itself cannot change and point to a different memory location. In other words, the pointer is constant, but the data at which it points is not.
• A pointer to const means that the pointer can point to different locations, but the data at the memory location the pointer is pointing to cannot be changed using this pointer. In other words, the data is constant but the pointer is not.
• A const pointer to const combines both concepts – the pointer cannot point to a different memory location and the data at the memory location cannot be changed. Both the pointer and the data it points to are constant.

In this lecture, we learned about Const pointer and Pointer to Const and Const Pointer to Const in C++.

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On compiling the program below the error "assignment of read-only location '1107307520u->COUNT16.INTFLAG.bit.OVF' " appears

The TC_INTFLAG register is not write protected

Any good ideas?

The processor is a SAMD21G18A. THe compile error does not appear when using a similar statement with another TC3 register

 #include "sam.h "

 void TC3_Handler(void) {   if (TC3->COUNT16.INTFLAG.bit.OVF) // TC3 overflow interrupt?   { TC3->COUNT16.INTFLAG.bit.OVF = 1u;    /* Clear the interrupt flag. This line causes the error */      TC3->COUNT16.INTFLAG.reg |= TC_INTFLAG_OVF; // No compile error!   } }

int main(void) {     /* Initialize the SAM system */     SystemInit();

    /* Replace with your application code */     while (1)      {     } }

The TC_INTFLAG_Type typedef in tc.h has this comment

So this is about avoiding a read-modify-write (that is the likely implementation of bit assignment) that would cause all flags that are currently 1 to be cleared (instead of just clearing OVF in your case).

A read-modify-write (and thus clearing all set flags) is what you have in the compiling statement:

better use:

But obviously that only makes a difference if you are actually using other flags in the register.

Thanks again for the very useful and illuminating answer - I see that I'm still on the SAMD learning curve, the AT(x)mega MCUs clear the flag when they jump to the ISR

At the moment I'm about 400km away from the hardware, so I'll have to wait for a test.

On p. 670 of the SAMD21 datasheet it says that writing a 0 to an INTFLAG bit has no effect, so if other bits are set your suggested code:

will leave them set.

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Home » C programs » C common errors programs

Error: Assignment of read-only variable in C | Common C program Errors

Here, we will learn how and when error 'assignment of read-only variable in C' occurs and how to fix it? By IncludeHelp Last updated : March 10, 2024

Error: Assignment of read-only variable in C

Error "assignment of read-only variable in C" occurs, when we try to assign a value to the read-only variable i.e. constant.

In this program, a is a read-only variable or we can say a is an integer constant, there are two mistakes that we have made:

• While declaring a constant, value must be assigned – which is not assigned.
• We cannot assign any value after the declaration statement to a constant – which we are trying to assign.

Consider the program:

How to fix it?

Assign value to the variable while declaring the constant and do not reassign the variable.

Correct Code

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• Error: expected '=', ',', ',' 'asm' or ' _attribute_' before '
• Error: Id returned 1 exit status (undefined reference to 'main')
• Error: Assignment of read-only location in C

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assignment of read-only location (vectors)

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Hey guys, I just finished semester one at school, and I have been doing some coding exercises to keep sharp over the winter break, and I have been some new errors on this exercise. This exercise came from a "top-coder" practice problem that was used a few years ago. Here is the problem statement: Code: ........ A chessboard pattern is a pattern that satisfies the following conditions: The pattern has a rectangular shape. The pattern contains only the characters '.' (a dot) and 'X' (an uppercase letter X). No two symbols that are horizontally or vertically adjacent are the same. The symbol in the lower left corner of the pattern is '.' (a dot). You are given two ints rows and columns. Write a method that computes the chessboard pattern with these dimensions, and returns it in a vector <string>. The elements of the return value correspond to rows of the pattern. Specifically, the first character of the last element of the return value represents the lower left corner I addressed the problem by writing this function here: Code: vector <string> makeChessboard (int w, int h){ string row = ""; vector<string>Board(); int i = 0; for(int height = 0; height < h; height++){ while(i<(w*h)){ for(int j = 0; j<w; j++){ if (i % 2 == 0){ row[j] = '.'; } else{ row[j] = 'X'; } i++; } } Board[height] = row; } return Board; } I am getting many errors on line 23 were I say Code: Board[height] = row; , including: Code: |23|warning: pointer to a function used in arithmetic |23|error: assignment of read-only location '*(Board + ((unsigned int)height))'| |23|error: cannot convert 'std::string' to 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >()' in assignment| |25|error: conversion from 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > > (*)()' to non-scalar type 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >' requested| I have no clue why I am getting those errors, I have never ran into the first warning before, could someone please shed some light on what I am doing wrong? I tried using push back also to no avail. I don't know why it is complaining about trying to convert between types. If you guys could help out with this I would appreciate it a ton! Thanks a ton guys! I appreciate it!

Last edited by jlangfo5; 12-16-2010 at 09:30 PM .

Code: vector<string>Board(); Here you declare Board to be a function returning a vector<string> (much like the function you're in, as a matter of fact), not an object of that type. If you want an empty constructor list, then simply leave off the parentheses.

Well thanks tabstop! I was able to fix that and it then compiled, but then everyone's favorite, the segfault appeared. I tried to test my function to see if it would work correctly, with this code here. Code: #include <iostream> #include <vector> using namespace std; vector <string> makeChessboard (int w, int h){ string row = ""; vector<string>Board; int i = 0; for(int height = 0; height < h; height++){ while(i<(w*h)){ for(int j = 0; j<w; j++){ if (i % 2 == 0){ row[j] = '.'; } else{ row[j] = 'X'; } i++; } } Board[height] = row; } return Board; } int main() { int h = 8; int w = 8; vector<string>chess; cout<< "making chess"; chess = makeChessboard (h, w); cout << "chess made"; for(int i = 0; i<chess.size(); i++){ cout << chess[i] << endl;} return 0; } It outputs "making chess", but crashes when it tries to do the assignment. So, it never gets to "chess made". Any takers? Thanks once again guys!

You are using the vector without first resizing it to appropriate size. Also, you forgot to include <string>. Same thing with the string. You are trying to access individual characters when it's empty. Logic is also wrong. You seem to be doing it in a overcomplicated way. I'd sit down and think a little more about how to accomplish it.

Last edited by Elysia; 12-17-2010 at 03:55 AM .

Originally Posted by Adak io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions. Originally Posted by Salem You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much. Outside of your DOS world, your header file is meaningless.

Thanks Elysia! It compiled after I used push back on the string instead of trying to assign to it like i was! Now I just have to fix a logic that error that isn't so bad, and it will be working properly! I'll post it when i have it working right!

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  Readonly class variables in C++

ValueError: assignment destination is read-only [Solved]

Last updated: Apr 11, 2024 Reading time · 2 min

# ValueError: assignment destination is read-only [Solved]

The NumPy "ValueError: assignment destination is read-only" occurs when you try to assign a value to a read-only array.

To solve the error, create a copy of the read-only array and modify the copy.

You can use the flags attribute to check if the array is WRITABLE .

Running the code sample produces the following output:

In your case, WRITEABLE will likely be set to False .

In older NumPy versions, you used to be able to set the flag to true by calling the setflags() method.

However, setting the WRITEABLE flag to True ( 1 ) will likely fail if the OWNDATA flag is set to False .

You will likely get the following error:

• "ValueError: cannot set WRITEABLE flag to True of this array"

To solve the error, create a copy of the array when converting it from a Pillow Image to a NumPy array.

Passing the Pillow Image to the numpy.array() method creates a copy of the array.

You can also explicitly call the copy() method.

You can also call the copy() method on the Pillow Image and modify the copy.

The image copy isn't read-only and allows assignment.

You can change the img_copy variable without getting the "assignment destination is read-only" error.

You can also use the numpy.copy() method.

The numpy.copy() method returns an array copy of the given object.

The only argument we passed to the method is the image.

You can safely modify the img_copy variable without running into issues.

You most likely don't want to make changes to the original image.

Creating a copy and modifying the copy should be your preferred approach.

If you got the error when using the np.asarray() method, try changing it to np.array() .

Change the following:

To the following:

As long as the WRITEABLE flag is set to True , you will be able to modify the array.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• TypeError: Object of type ndarray is not JSON serializable
• ValueError: numpy.ndarray size changed, may indicate binary incompatibility
• NumPy RuntimeWarning: divide by zero encountered in log10
• ValueError: x and y must have same first dimension, but have shapes

Borislav Hadzhiev

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Copyright © 2024 Borislav Hadzhiev

Constant pointer and pointer to constant in C

Quick links.

• Constant pointer
• Pointer to constant
• Constant pointer to constant

Pointers are the most powerful as well as complex component of C programming. For newbies, it’s like learning rocket science in C. However, I have tried my best to simplify things.

In this ongoing series of C programming tutorial, I have explained many concepts related to pointers. Here in this section we will focus on some confusing pointer terminologies. We will learn and compare constant pointer with pointer to constant and constant pointer to constant .

What is constant pointer?

To understand a constant pointer, let us recall definition of a constant variable . Constant variable is a variable whose value cannot be altered throughout the program.

Similarly, constant pointer is a pointer variable whose value cannot be altered throughout the program. It does not allows modification of its value, however you can modify the value pointed by a pointer.

In other words, constant pointer is a pointer that can only point to single object throughout the program.

Syntax to declare constant pointer

Note: You must initialize a constant pointer at the time of its declaration.

Example to declare constant pointer

Note: We use const keyword to declare a constant pointer.

The compiler will generate compilation error on failure of any of the two conditions.

• You must initialize a constant pointer during its declaration.
• A constant pointer must not be re-assigned.

Example program to use constant pointer

Let us write an example program to demonstrate constant pointer in C.

Let us dry run the above program.

• First, we declared two integer variable num1 , num2 and an integer constant pointer const_ptr that points to num1 .
• The statement *const_ptr = 10; assigns 10 to num1 .
• Next we tried re-assignment of constant pointer i.e. const_ptr = &num2; . The statement will generate compilation error, since a constant pointer can only point to single object throughout the program.
• The last two printf() statements are used to test value of num1 .

On compilation it generates following error message.

What is pointer to constant?

Pointer to constant is a pointer that restricts modification of value pointed by the pointer. You can modify pointer value, but you cannot modify the value pointed by pointer.

Note: There is a minor difference between constant pointer and pointer to constant. A constant pointer can only point to single object throughout the program. You can, alter the value pointed by pointer, but cannot alter pointer value. Whereas pointer to a constant cannot modify the value pointed by pointer, but can alter its value.

Syntax to declare pointer to constant

Example to declare pointer to constant, example program to use pointer to constant.

Let us demonstrate pointer to a constant using an example.

The above program generates compilation error, since we tried to assign value to read only memory location (i.e. the value pointed by the pointer). Below is the compilation error generated by *ptr_const = 100;

Note: Pointer to constant restricts modification of value pointed by the pointer. However, do not think that C compiler converts variable pointed by pointer as constant variable.

Pointer to constant does not allows you to modify the pointed value, using pointer. However, you can directly perform modification on variable (without using pointer). For example,

What is constant pointer to constant?

A constant pointer to constant is a combination of constant pointer and pointer to constant . It is a pointer that does not allow modification of pointer value as well as value pointed by the pointer.

Syntax to declare constant pointer to constant

Example to declare constant pointer to constant, example program to use constant pointer to constant.

Let us demonstrate the concept of constant pointer to constant in C program.

The above program generates two compilation error. First in the statement ptr = &num2; since we tried to assign value to a constant pointer. Second in the statement *ptr = 100; since we tried to assign value pointed by a pointer to constant.

Difference between constant pointer, pointer to constant and constant pointer to constant

Okay so got basic picture of a constant pointer, pointer to constant and constant pointer to constant. Let us summarize the things for quick recap.

【C言語】assignment of read-only location ‘xxx’

“assignment of read-only location ‘xxx'” というエラーメッセージは、読み取り専用の場所に代入しようとした場合に表示されます。

例えば、次のようなコードを書いたとします。

このコードでは、定数 x の値を 10 から 20 に変更しようとしています。しかし、定数は値を変更することができないため、このコードは “assignment of read-only location ‘x'” というエラーを引き起こします。

正しいコードは、次のようになります。

この場合、定数 x の値を変数 y にコピーし、変数 y の値を変更しています。そのため、このコードはエラーを引き起こしません。

• unordered-map

"error: assignment of read-only location" in unordered_map (C++)

I have an awkward hash table (specifically, an unordered_map) with int keys and vector< vector< int >> data. I periodically need to update elements in this two-dimensional vector of ints. There's no intrinsic reason I shouldn't be able to, right? A newer g++ compiler I've switched to complains of an assignment of read-only location on the line designated below.

I'm new to C++, so nothing's too obvious. Thank you for any help.

Following Tim's suggestion

I've replaced the relevant parts of code above with the following:

This code compiles without the error and appears to run fine. Like Tim, I still don't quite understand why the fix works. The error was previously appearing with gcc version 4.1.2 20080704 (Red Hat 4.1.2-44) but not with gcc version 4.0.1 (Apple Inc. build 5465). I will try to dissect the error more carefully when I'm not under a tight deadline!

Are you sure there really are thisStep + 1 elements in every first-level vector and NUM_DEMES elements in every second-level vector?

You're not actually assigning to a map iterator, if I read correctly, so I suspect the error is in the vector access.

It may be helpful to break that last statement up into multiple statements so that only one thing is being done on each to narrow down where the problem is. e.g.,

By the way, piItr = phenotypeIs.begin(); has no effect here, it could be simply:

at() returns an iterator into the inner vector, not access to the value. What you want is

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