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
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