I tried assigning a signed int to unsigned int.
#include <stdio.h>
int main()
{
int a;
unsigned int b;
scanf("%d", &a);
b = a;
printf("%d %u\n", a, b);
return 0;
}
I was hoping that compiling this would cause a warning that I am assigning an int value to unsigned int variable. But I did not get any warning.
$ gcc -std=c99 -Wall -Wextra -pedantic foo.c
$ echo -1 | ./a.out
-1 4294967295
Next I tried to assigning an unsigned int to signed int.
#include <stdio.h>
int main()
{
int a;
unsigned int b;
scanf("%u", &b);
a = b;
printf("%d %u\n", a, b);
return 0;
}
Still no warning.
$ gcc -std=c99 -Wall -Wextra -pedantic bar.c
$ echo 4294967295 | ./a.out
-1 4294967295
Two questions:
Code 1: This conversion is well-defined. If the int is out of range of unsigned int, then UINT_MAX + 1 is added to bring it in range.
Since the code is correct and normal, there should be no warning. However you could try the gcc switch -Wconversion which does produce a warning for some correct conversions, particularly signed-unsigned conversion.
Code 2: This conversion is implementation-defined if the input is larger than INT_MAX. Most likely the implementation you are on defines it to be the inverse of the conversion in Code 1.
Typically, compilers don't warn for implementation-defined code which is well-defined on that implementation. Again you can use -Wconversion.
A cast is not necessary and as a general principle, casts should be avoided as they can hide error messages.
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