CODEWITHSUNDEEP
c
Anish Ramaswamy
Anish Ramaswamy

Reputation: 2341

Why does this program result in an overflow?

#include <stdio.h>
int main(void)
{
    char test = 0x80;
    printf("%c\n", test);  /* To remove the "unused" warning */
    return 0;
}

I understand that a character is guaranteed to be one byte. 0x80 is also one byte. So why then do I get the following error/warning?

error: overflow in implicit constant conversion [-Werror=overflow]

In my case it's an error because I'm treating warnings as errors.

0x80 is the minimum value for which this warning/error appears. If I change it to 0x7F, this compiles fine. I used ideone.com with the 'C99 strict' option to compile the code. It reported using gcc-4.7.2.

Upvotes: 3

Views: 744

Answers (2)

user2135588
user2135588

Reputation: 21

The variable test is defined as char which defaults to signed char. range of which is 0 ~ 127 (7 bits). The last bit is reserved for sign. Maybe the error is due to trying to print a character with negative value(0x80 = -128).

Try with unsigned char for test and check if you get same error.

Upvotes: 2

Jon
Jon

Reputation: 437336

The char type is signed in your compiler, so even though it does have 8 bits of information it cannot store values greater than 127 (0x7f).

The header <limits.h> defines macros that let you determine the signedness and range limits for integral types, including char.

Upvotes: 13

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