Is this undefined behaviour?

Code:

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#include <stdio.h>

int main(void)
{
    char v[] = "blablabla";

    v[0] = 'b';

    return 0;
}

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I know this is, but what about this?

Code:

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#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void)
{
    char *v = malloc(strlen("blablabla") + 1);
    if (v)
        strcpy(v, "blablabla");

    v[0] = 'b';

    free(v);

    return 0;
}

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I think it is ok but I just want to make sure. Thank you.

Neither is UB. You are thinking of the UB of writing to a string literal.

if char *v = "blablabla";
this one is undefined behaviour.

Code:

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The C99 Draft (N869, 18 January, 1999)
J.2 Undefined behavior
#1
The behavior is undefined in the following circumstances: ...
An attempt is made to modify a string literal of either form (6.4.5).

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EDIT=edited.thanks xeddiex and eerok.

Code:

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  char *s1 = "string literal";
  char s2[] = "not a string literal";

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Quote:

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Originally Posted by eerok

Code:

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  char *s1 = "string literal";
  char s2[] = "not a string literal";

---

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immutable: char *s1
They're both string literals but the difference is that in the first one, s1 is being assigned a literal that is located in the data segment (always?) where all your programs execution instructions are stored along with other life time program duration stufff and you cannot and shouldn't try to change anything there.

mutable: char s2[]
The latter one has two stages; first the memory for s2 is allocated on the stack and then the string literal (which is located in the data segment) on the right is copied character for character to each block in memory in s2. And the values in s2 are now different from the string literal because they're both different copies, so, it's safe to write to.