Looping char* gives me Access violation writing location 0x00CFB310

Question

i try to print simple c* string like this :

char *cc = "HEllo";

                for (char* inputPtr = cc; inputPtr[0]; inputPtr++) {
                    char c = inputPtr[0]++;
                    printf("%s",c);
                }

but im getting : Access violation writing location 0x00CFB310. on :

char c = inputPtr[0]++;

what is wrong here ?

@dandan78 This is not a good duplicate, because OP is not realizing he's incrementing the char, not the pointer. — Sergey Kalinichenko, Mar 04 '17 at 10:15

Sergey Kalinichenko · Accepted Answer · 2017-03-04T10:01:15.450

2

It looks like you are experimenting with inputPtr[0] as a substitution for *inputPtr. In many contexts, the two expressions produce the same result.

However, expression inputPtr[0]++ is not the same as *inputPtr++, because [0] has higher precedence than *, but it has the same precedence as suffix ++. Operators within this precedence level are applied left-to-right, so the first expression post-increments inputPtr[0], a character inside a character literal. This is undefined behavior, hence you see a crash.

If you replace inputPtr[0]++ with *inputPtr++ and remove inputPtr++ from loop header, your code is going to work fine:

for (char* inputPtr = cc; inputPtr[0]; ) {
    char c = *inputPtr++;
    printf("%c", c); // Replace %s with %c to print one character
}

edited Mar 04 '17 at 10:01

answered Mar 04 '17 at 09:52

Sergey Kalinichenko

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The code with `inputPtr[0]++` changed to `*inputPtr++` will invoke *undefined behavior* for reading out-of-range. `inputPtr` is now incremented twice per loop and the terminating null-character will be skipped. – MikeCAT Mar 04 '17 at 09:56
@MikeCAT You're right, he's doing a `for` loop, not a `while` loop, where it would make more sense. Thanks! – Sergey Kalinichenko Mar 04 '17 at 09:58
Im getting Access violation reading location 0x00000048. on the printf – user63898 Mar 04 '17 at 10:00
1

@user63898 That's because you used `%s` in place of `%c`. Edited. – Sergey Kalinichenko Mar 04 '17 at 10:01

MikeCAT · Answer 2 · 2017-03-04T10:04:30.990

1

inputPtr is pointing at "HEllo", which is a string literal.

Modifying string literal isn't allowed and trying to do so invokes undefined behavior.

inputPtr[0]++ is trying to modify string literal. If data of string literal is located in read-only locations, it may lead to Segmentation Fault.

edited Mar 04 '17 at 10:04

answered Mar 04 '17 at 09:50

MikeCAT

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you could suggest fix `char cc[] = "HEllo";` then – Jean-François Fabre Mar 04 '17 at 09:52
@Jean-FrançoisFabre why, when `inputPtr[0] + 1` would have the same effect here. – Antti Haapala -- Слава Україні Mar 04 '17 at 09:53
good point. I made some selective reading. Why on earth modifying some input data in-place? someone should suggest that to the OP instead of just pointing out the error (which makes that a duplicate question) – Jean-François Fabre Mar 04 '17 at 09:55
@AnttiHaapala I don't think `inputPtr[0] + 1` will have the same effect because post-increment is used in the original code and `inputPtr[0]`, not `inputPtr[0] + 1` will be stored to `c` if `inputPtr[0]` were writable. – MikeCAT Mar 04 '17 at 10:05

score 0 · Answer 3 · answered Mar 04 '17 at 09:51

0

String literals in C are read only, attempting to modify the characters in string literals leads to undefined behavior.

Any with inputPtr[0] you do modify the string since you increment the character inputPtr[0].

This is the reason you should always use const char * when pointing to string literals.

If you want to modify the contents of the string you have to create an array:

char cc[] = "HEllo";

answered Mar 04 '17 at 09:51

Some programmer dude

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this char pointer in reallty im getting from API call , so how do i loop it ? do i need to make copy of it to char[]? – user63898 Mar 04 '17 at 09:55
@user63898 If you want to modify the contents of the string, then yes you should definitely create a copy. – Some programmer dude Mar 04 '17 at 10:31

score 0 · Answer 4 · answered Mar 04 '17 at 09:52

0

As the others have written, you can't modify a string literal. Change your declaration to char cc[] = "HEllo"; and see what happens. The suggested declaration declares a string buffer that is modifiable.

answered Mar 04 '17 at 09:52

lukeg

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Looping char* gives me Access violation writing location 0x00CFB310

4 Answers4