Question: What will happen when the following code runs? char * c = malloc ( 1 ) ; c [ 0 ] = ' d '

What will happen when the following code runs?
char *c = malloc(1);
c[0]='d';
Group of answer choices
The value at address c is set to 'd'.
An error occurs.
c points to 'd' which is at a different address from what was returned from malloc.

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