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When can I not declare an int with a signature?

In C, it is assumed that declared integers like -1 are declared with a keyword signed, for example:

signed int i = -1;

However, I tried this:

signed int i = -2;
unsigned int i = -2;
int i = -2;

and all 3 cases print -2 with printf("%d", i);. Why?

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5 answers

Since you have confirmed that you are printing using:

printf("%d", i);

this is undefined behavior in an unsigned case. This is described in the C99 draft section of the 7.19.6.1fprintf function, which also covers printfformat specifiers, as stated in clause 9:

If the conversion specification is not valid, the behavior is undefined. 248) [...]

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unsigned int i = -2; // i actually holds 4294967294
printf("%d", i); // printf casts i back to an int which is -2 hence the same output
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