Why is sizeof(int) less than
This question already has an answer here:
sizeof generates a size_t which is always positive. You are comparing it with -1 which is probably promoted in size_t which gave you a HUGE number, most likely greater than the size of an int.
To make sure of it, try this:
printf("%zun", sizeof(int));
printf("%zun", (size_t)(-1));
[EDIT]: Following comments (some have been removed), I precise indeed that sizeof is an operator, not a function.
From the standard, C11, 6.5.3.4 The sizeof and _Alignof operators :
5 The value of the result of both operators is implementation-defined, and its type (an unsigned integer type) is size_t, defined in (and other headers).
So, the sizeof operator yields an unsigned value. You then compare an unsigned value with a signed value. That is dealt with by converting the signed value to be unsigned, which explains the behaviour that you see.
First, running your code:
cc1plus: warnings being treated as errors
In function 'int main()':
Line 3: warning: comparison between signed and unsigned integer expressions
Then fixing it:
int main()
{
if ((int)sizeof(int) > -1) // cast value of sizeof(int) to int before compare to int
printf("True");
else
printf("False");
return 0;
}
Result:
True
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