Strtol doesn't set errno on overflow conversion

Question

My strtol function fails to set errno during overflown conversion.

#include <stdio.h>
#include <string.h>
#include <malloc.h>
#include <getopt.h>
#include <errno.h>
#include <stdlib.h>    

int main(int argc, char **argv) {

    errno = 0;
    int e = strtol("1000000000000000", NULL, 10);
    printf("%d %d\n", errno, e);

    return 0;
}

returns

0 -1530494976

What do I do wrong?

Compiler

gcc (Ubuntu 4.9.2-10ubuntu13) 4.9.2

Options

gcc -Wall -std=gnu99 -O2

What compiler (and its version) are you using? Did you include all the necessary headers? — P.P, Aug 19 '15 at 20:16
`int` coercion happens outside of `strtol`, so `strtol` can't detect range error anyway. — fukanchik, Aug 19 '15 at 20:32
Example [`strtoi()`](http://stackoverflow.com/a/29378380/2410359) may help. — chux - Reinstate Monica, Aug 19 '15 at 20:37

cremno · Accepted Answer · 2015-08-19T21:01:53.863

5

There is nothing wrong with the implementation of strtol() but there is with your code.

The return type of this function is long (see the trailing l) and apparently the value 1000000000000000 can be represented by the long integer type. However the return value is assigned to e whose type is int which is unable to represent this value. What then happens is implementation-defined.

So change int e to long e and "%d %d\n" to "%d %ld\n". If you want to keep it as int, then you have to check if the value is outside of its range of representable values by yourself:

#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <limits.h>  // for INT_{MIN,MAX}
int
main(void)
{
    errno = 0;
    long f = strtol("1000000000000000", NULL, 10);
    if (errno == ERANGE) {
        puts("value not representable by long (or int)");
    } else if (f < INT_MIN || f > INT_MAX) {
        puts("value not representable by int");
    } else {
        int e = f;
        printf("%d\n", e);
    }
}

edited Aug 19 '15 at 21:01

answered Aug 19 '15 at 20:25

cremno

4,672
1
16
27

Thanks for the answer. Do you possibly know if there is a function for normal int conversion, with overflow detection? – Krzysztofik Aug 19 '15 at 20:28
@Krzysztofik: I've added an example. – cremno Aug 19 '15 at 20:42
@cremno Actually, it was a brain fart. You're right, this is LP64, not LLP64. – Sneftel Aug 19 '15 at 20:42
@cremno Isn't `long` equal in size to `int`? Tested your code, see http://ideone.com/awPacv. – Philip Aug 19 '15 at 20:48
@Philip Dahnen `long` has _at least_ the range of `int`. There are the same on many platform, yet differ on others - as allowed by the C spec. – chux - Reinstate Monica Aug 19 '15 at 20:51

score -1 · Answer 2 · answered Aug 19 '15 at 20:24

-1

It seems like both Microsoft [1] and Apple [2] implementations have the setting of errno commented out.

[1] http://research.microsoft.com/en-us/um/redmond/projects/invisible/src/crt/strtol.c.htm

[2] http://www.opensource.apple.com/source/xnu/xnu-1456.1.26/bsd/libkern/strtol.c

answered Aug 19 '15 at 20:24

Philip

271
1
15

He's using GCC on Ubuntu. – Barmar Aug 19 '15 at 20:30
1

`strtol` can't detect `int` coercion which happens outside of `strtol` – fukanchik Aug 19 '15 at 20:30
1

It's also not the Visual C++ CRT source code and the Apple link looks like kernel and not libc code. – cremno Aug 19 '15 at 20:42

Strtol doesn't set errno on overflow conversion

2 Answers2