I noticed that something happens when posting my message and it got messed. Reposting: The warning about overflows in int conversions show up only when std=99 option is used. In non-c99 mode, a different warning shows up. However, warning about char overflow shows up in both. Examples cat << EOF | gcc -c -xc -pedantic -std=c99 - > char foo = 255; > signed int ia = 2166136261; > signed int ia1 = 2147483648; > EOF <stdin>:1:1: warning: overflow in implicit constant conversion [-Woverflow] <stdin>:2:1: warning: overflow in implicit constant conversion [-Woverflow] <stdin>:3:1: warning: overflow in implicit constant conversion [-Woverflow] cat << EOF | gcc -c -xc -pedantic - > char foo = 255; > signed int ia = 2166136261; > signed int ia1 = 2147483648; > EOF <stdin>:1:1: warning: overflow in implicit constant conversion [-Woverflow] <stdin>:2:1: warning: this decimal constant is unsigned only in ISO C90 [enabled by default] <stdin>:3:1: warning: this decimal constant is unsigned only in ISO C90 [enabled by default] It looks like in case of int gcc simply compares if lvalue has the same size as rvalue. Integer constants conversion rules have changed between C90 and C99. In C90, 2166136261 is unsigned long int compiler which has size of 4, the same as int hence no warning. But in C99 2166136261 is long long int which is 8 bytes. OTOH, if `u' suffix is add to integer constant making it unsigned integer no warning is generated in both modes: cat << EOF | gcc -c -xc -pedantic - > signed int ia = 2166136261u; > signed int ia1 = 2147483648u; > EOF cat << EOF | gcc -c -xc -pedantic -std=c99 - > signed int ia = 2166136261u; > signed int ia1 = 2147483648u; > EOF
