strtoul() not working as expected?

Question

I am trying to convert a string such as "0x7ffd01767a60" to hexadecimal so I can compare pointers. Not sure if this is the best decision.

I am doing this:

    char *address = "0x7ffd01767a60";

    strtol(address,NULL,16);
    printf("%lp",address);

And I am getting this: 0x7ffd01764120

EDIT: It seems I was printing the string address ignoring the function return. Thanks Jens! and schlenk.

SOLVED! This is what I do

    char *address = "0x7ffd01767a60";
    void *p;
    unsigned long int address_hex = strtol(address,NULL,16);
    p = (void*) address_hex;

    printf("%p",p);

printf prints the same memory address.

Where do you capture the result of [strtoul](http://linux.die.net/man/3/strtoul)? All you do here is print the pointer to the string `"0x7ffd01767a60"` which is different from the content of the string. — Jens, Nov 28 '15 at 00:02
What do you expect this to do? `strtoul()` has a return value which you simply ignore. — schlenk, Nov 28 '15 at 00:02
@deko: If the answer below satisfies your question, please consider accepting it. — Jens, Dec 02 '15 at 04:17

Jens · Accepted Answer · 2015-12-03T05:07:53.977

3

You're printing the address of the string itself while ignoring the result of the strtoul() function call. This should work:

const char *address = "0x7ffd01767a60";

unsigned long int address_hex = strtoul(address, NULL, 16);
// Check for errors in errno
printf("%lx\n", address_hex);

Also, personally I prefer code to be as explicit as possible which is why I passed 16 as the base parameter.

Note: Please read the documentation on the return value, to make sure that you identify errors correctly.

edited Dec 03 '15 at 05:07

answered Nov 28 '15 at 00:08

Jens

8,423
9
58
78

hey Jens, it won't let me to print that, shouldn't it be %lu instead of %p since it is a long int? using %lu I get: 140733653415520 – deko Nov 28 '15 at 00:15
@deko: Correct, my bad. According to the [printf()](http://linux.die.net/man/3/printf) documentation, the type for `%p` is a `void *` pointer type, but `address_hex` is an integer. Using `%lx` would be correct in your case assuming you're looking for a hex printout; `%lu` results in a decimal. (Also, I'd avoid casting the integer to a pointer because you might run into problems on architectures where your pointer width is different from your (long) integer width.) – Jens Nov 28 '15 at 04:47

strtoul() not working as expected?

1 Answers1