C++ long overflowing prematurely
I'm having a bizarre problem with C++ where the long data type is overflowing long before it should. What I'm doing (with success so far) is to have integers behave like floats, so that the range [-32767,32767] is mapped to [-1.0,1.0]. Where it stumbles is with larger arguments representing floats greater than 1.0:
inline long times(long a, long b) {
printf("a=%ld b=%ld ",a,b);
a *= b;
printf("a*b=%ld ",a);
a /= 32767l;
printf("a*b/32767=%ld\n",a);
return a;
}
int main(void) {
printf("%ld\n",times(98301l,32767l));
}
What I get as output is:
a=98301 b=32767 a开发者_开发问答*b=-1073938429 a*b/32767=-32775
-32775
So times(98301,32767) is analogous to 3.0*1.0. This code works perfectly when the arguments to times are less than 32767 (1.0), but none of the intermediate steps with the arguments above should overflow the 64 bits of long.
Any ideas?
long is not necessarily 64 bits. try 'long long' instead.
The type long
is not necessarily 64 bits. If you are on the 32 bit architecture (at least on MS Visual c++), the long
type is 32 bits. Check it out with sizeof (long)
. There is also the long long
data type that may help.
You probably have 32-bit longs. Try using long long
instead.
98301 * 32767 = 3221028867, while a 32-bit long overflows at 2147483648
The C standard only guarantees that long
will have at least 32 bit (which is actually the case on most 32bit platforms).
If you need 64 bit, use long long
. It's guaranteed to hold at least 64 bit.
精彩评论