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Asked: 2026-06-08T15:24:23+00:00

Let’s say I have 2 functions: void function1(int *ptr) { printf(%d, *ptr); } and

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Let’s say I have 2 functions:

void function1(int *ptr) {
     printf("%d", *ptr);
}

and

void function2(char *str) {
     printf("%s", str);
}

Why does function2 work, when there is no deference operator before str? In str there is only the address where it points to not the value as I thought.

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1 Answer

  1. Editorial Team

    Why does function2 work, when there is no deference operator before
    str

    Because %s, as defined by the standard, expects a char *, i.e. an address.

    7.21.6.1

    If no l length modifier is present, the argument shall be a pointer to
    the initial element of an array of character type    . Characters
    from the array are written up to (but not including) the
    terminating null character. If the precision is specified, no more
    than that many bytes are written. If the precision is not specified or is
    greater than the size of the array, the array shall contain a null character.

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