Change char by index

Question

Trying to change char by index:

char *p = "test";
for (int i = 0; i < 4; ++i) {
    p[i] = 50;
}
printf("%s\n", p);//test

What I'm doing wrong?

You’re trying to change a character literal which is read-only — Sami Kuhmonen, Feb 15 '22 at 22:21
@SamiKuhmonen: You probably mean ["string literal"](https://en.cppreference.com/w/c/language/string_literal), not "character literal". — Andreas Wenzel, Feb 15 '22 at 22:24
@AndreasWenzel Sorry, yes, that’s exactly what I meant of course — Sami Kuhmonen, Feb 16 '22 at 05:35

score 1 · Answer 1 · answered Feb 15 '22 at 22:25

1

you cannot change a literal,

char*p="test";

doesnt copy the literal. You should do

char p[] = "test";

or

#include <string.h>
....
char *p = strdup("test");

if you want to deal with a pointer rather than a char array.

answered Feb 15 '22 at 22:25

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score 0 · Answer 2 · answered Feb 15 '22 at 22:36

The line

char *p = "test";

will make the pointer p point to the string literal "test".

String literals are read-only. Attempting to modify them will invoke undefined behavior.

If you want "test" to be writable, then you must allocate (writable) memory for it, for example by declaring an array:

char arr[] = "test";

That line is equivalent to

char arr[5] = "test";

which creates an array of 5 characters and initializes the first 4 characters to "test" and initializes the 5^th character to a null terminating character.

Now that you have allocated memory for the string, you can write to it at will, for example you could write:

for (int i = 0; i < 4; ++i) {
    arr[i] = 50;
}

Or you could make a pointer p point to the array, and use the pointer for accessing the string, as you did in the code in the question:

char *p = arr;
for (int i = 0; i < 4; ++i) {
    p[i] = 50;
}

Change char by index

2 Answers2