What is the difference between strtol and strtoul?

Question

I met some unexcepted result of strtol in c

Here is the sample program.

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

int main()
{
    printf("%x\n", strtol("0xfffff70A", NULL, 0));
    return 0;
}

and the output of this simple program is

0x7fffffff

rather than 0xfffff70A. And if I use strtoul, the result is exactly 0xfffff70a. I am using a 32-bit machine, and I wonder what happens. PS. I am using gcc 4.7.2

Set `errno` to `0` before calling `strtol` or `strtoul` and check it afterwards. Do not use the value if `errno` has a different value. — pmg, May 23 '13 at 10:54
Nice question, by the way: Next to minimal code example, input, observed output, expected output. +1 for making this so easy to answer. — DevSolar, May 23 '13 at 11:14

score 7 · Accepted Answer · edited May 23 '13 at 11:05

7

From 7.22.1.4 paragraph 8 (of the N1570 draft of the 2011 edition of the standard):

If the correct value is outside the range of representable values, LONG_MIN, LONG_MAX, LLONG_MIN, LLONG_MAX, ULONG_MAX, or ULLONG_MAX is returned (according to the return type and sign of the value, if any), and the value of the macro ERANGE is stored in errno.

Since the correct value of your input string is too large for the type, you get LONG_MAX, and errno is set to ERANGE.

Whenever one of the strto(u)l(l) functions returns one of the TYPE_MAX or TYPE_MIN values, you need to check errno to find out whether it's a correct result, or your input was out-of-range.

edited May 23 '13 at 11:05

pmg

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answered May 23 '13 at 10:52

Daniel Fischer

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Damn, I was *just* about to paste the same. :-D @PascalCuoq: Behaviour of the `strto...` functions is very well-defined even in case of overflows. That's what makes them superior to e.g. `*scanf()` functions for parsing numerical input. – DevSolar May 23 '13 at 10:53
Can you also specify from where you're quoting? – Olaf Dietsche May 23 '13 at 10:55
@OlafDietsche: That would be ISO/IEC 9899 (the C language standard). – DevSolar May 23 '13 at 10:56
@DevSolar Not everybody knows that :-) – Olaf Dietsche May 23 '13 at 10:57
In contrast, why can `(int)strtoul("-2",NULL,0)` return the "correct" value? – user1668903 May 23 '13 at 11:38
@user1668903 From paragraph 5: "If the subject sequence begins with a minus sign, the value resulting from the conversion is negated (in the return type)." The `"2"` part of the input is parsed as the value 2 of type `unsigned long`. Then that value is negated, that makes it `ULONG_MAX + 1 - 2`, (the mathematical result is reduce modulo `ULONG_MAX + 1` to obtain a value in the range `[0, ULONG_MAX]`). Then that is converted to `int` in an implementation-defined manner (assuming `ULONG_MAX > INT_MAX`). Usually, the least significant bits are just reinterpreted, and in two's complement that's -2 – Daniel Fischer May 23 '13 at 11:50
@DevSolar You are referring to a mistake that was in my answer for only two minutes! I remembered and fixed it immediately. You are right, functions of this family have well-defined behavior as long as they are passed a well-formed string. – Pascal Cuoq May 23 '13 at 13:09
@PascalCuoq: I was only here for a minute or so. ;-) Actually the `strto*()` functions are well-defined even for non-well-formed string. Nevermind, I think this was answered to satisfaction. ;-) – DevSolar May 23 '13 at 13:25
@DevSolar Here is a use of `strtol()` that results in undefined behavior. The undefined behavior is caused by `s` not being a well-formed string. http://ideone.com/Qs2i7n – Pascal Cuoq May 23 '13 at 13:33
@PascalCuoq: Ah... OK, *that* kind of non-well-formed-ness. ;-) I bow out of the discussion, you found a way to break the function. ;-) – DevSolar May 23 '13 at 13:53

score 1 · Answer 2 · answered May 23 '13 at 10:53

You're running into overflow of the long type, which is signed.

You probably should use:

print("%lx\n", strtoul("0xfffff70a", NULL, 0));
                    ^
                    |
                 important!

instead, note the 'u' for "unsigned" (see manual page).

Also note that you can't print an unsigned long with plain %x, you need to qualify it as being bigger than int.

score 0 · Answer 3 · answered May 23 '13 at 10:49

0

Your architecture has a 32-bit long type. 0xfffff70A is not representable as a signed 32-bit long. errno should have been set to ERANGE.

In 2's complement, representable values for 32-bit signed integers range from -0x80000000 to 0x7fffffff.

answered May 23 '13 at 10:49

Pascal Cuoq

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What is the difference between strtol and strtoul?

3 Answers3