Why is sizeof(int) less than -1? [duplicate]

问题

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void main() { if(sizeof(int) > -1) printf(“true”); else printf(“false”); ; [duplicate] (3 answers)

sizeof() operator in if-statement (5 answers)

Closed 5 years ago.

In the following code

#include <stdio.h>

int main()
{
   if (sizeof(int) > -1)
       printf("True");
   else
       printf("False");
return 0;
}

I get the output as "False" rather than "True".My understanding is that sizeof operator just returns the size of int which will be 4 in this case.

Why is the condition evaluated as false?

回答1:

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("%zu\n", sizeof(int));
printf("%zu\n", (size_t)(-1));

[EDIT]: Following comments (some have been removed), I precise indeed that sizeof is an operator, not a function.

回答2:

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.

回答3:

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

来源：https://stackoverflow.com/questions/24797843/why-is-sizeofint-less-than-1

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