Why is the integer not incremented in this code?

Question

Can anyone explain what I'm doing wrong here to not get 11 as my output?

void foo {
    int *n = malloc(sizeof(int)); 
    *n = 10; 
    n++;
    printf("%d", *n)
}

Answer 1

If we call the malloc'ed variable x, then your program does this:

                                      n     x
int *n = malloc(sizeof(int));        &x     ?
*n = 10;                             &x    10
n++;                                &x+1   10

You want to do this:

                                      n     x
int *n = malloc(sizeof(int));        &x     ?
*n = 10;                             &x    10
(*n)++;                              &x    11

Answer 2

n++ increments the pointer n, not the integer pointed to by n. To increment the integer, you need to dereference the pointer and then increment the result of that:

(*n)++;

Answer 3

You set n[0] to 10, and then you print n[1]. malloc() does not initialize the memory that it gives you, so what gets printed is unpredictable - it's whatever garbage happened to be in n[1].

Answer 4

You can get 11 as your output with this code:

void foo {
    int *n = malloc(sizeof(int)); 
    *n = 10; 
    (*n)++; 
    printf("%d", *n)
}

Answer 5

n++ moves the pointer sizeof(int) bytes forward.