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strtoull and long long arithmetic

开发者 https://www.devze.com 2022-12-09 15:52 出处:网络
Can anyone explain the output of this program and how I can fix it? unsigned long long ns = strtoull(\"123110724001300\", (开发者_如何学Pythonchar **)NULL, 10);

Can anyone explain the output of this program and how I can fix it?

unsigned long long ns = strtoull("123110724001300", (开发者_如何学Pythonchar **)NULL, 10);
fprintf(stderr, "%llu\n", ns);

// 18446744073490980372


Do you have <stdlib.h> included?

I can reproduce on MacOS X if I omit <stdlib.h>.

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

int main(void)
{
    unsigned long long ns = strtoll("123110724001300", (char **)NULL, 10);
    printf("%llu\n", ns);
    return(0);
}

Omit the header, I get your result. Include the header, I get the correct answer.

Both 32-bit and 64-bit compiles.


As noted in the comments, in the absence of a declaration for strtoll(), the compiler treats it as a function returning int.

To see more of what goes on, look at the hex outputs:

     123110724001300    0x00006FF7_F2F8DE14    Correct
18446744073490980372    0xFFFFFFFF_F2F8DE14    Incorrect

Manually inserted underscores...


Why not use strtoull if you want an unsigned long long?


I cannot explain the behavior. However, on 32 bit Windows XP with Cygwin gcc-4.3.2:

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

int main(void) {
    unsigned long long ns = strtoull("123110724001300", NULL, 10);
    fprintf(stderr, "%llu\n", ns);
    return 0;
}

prints

E:\Home> t.exe
123110724001300
0

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