Modifying a char array inside a function in C [duplicate]

Question

So I've been playing around with C lately and have been trying to understand the intricacies of passing by value/reference and the ability to manipulate a passed-in variable inside a function. I've hit a road block, however, with the following:

void modifyCharArray(char *input)
{
    //change input[0] to 'D'
    input[0] = 'D';
}

int main()
{
    char *test = "Bad";
    modifyCharArray(test);
    printf("Bad --> %s\n", test);
}

So the idea was to just modify a char array inside a function, and then print out said array after the modification completed. However, this fails, since all I'm doing is modifying the value of input that is passed in and not the actual memory address.

In short, is there any way I can take in a char *input into a function and modify its original memory address without using something like memcpy from string.h?

Answer 1

In short, is there any way I can take in a char *input into a function and modify its original memory address without using something like memcpy from string.h?

Yes, you can. Your function modifyCharArray is doing the right thing. What you are seeing is caused by that fact that

char *test = "Bad";

creates "Bad" in read only memory of the program and test points to that memory. Changing it is cause for undefined behavior.

If you want to create a modifiable string, use:

char test[] = "Bad";