CODEWITHSUNDEEP
c++overflowinteger-overflow
rhodri
rhodri

Reputation: 370

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*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?

Upvotes: 7

Views: 424

Answers (4)

sleske
sleske

Reputation: 83645

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.

Upvotes: 2

Kerido
Kerido

Reputation: 2940

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.

Upvotes: 4

Gabe
Gabe

Reputation: 86848

You probably have 32-bit longs. Try using long long instead.

98301 * 32767 = 3221028867, while a 32-bit long overflows at 2147483648

Upvotes: 2

Gregor Brandt
Gregor Brandt

Reputation: 7799

long is not necessarily 64 bits. try 'long long' instead.

Upvotes: 9

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