output - arithmetic overflow because signed characters in C program -

we know signed char can have values -128 127. when run below program no overflow happens though l output exceeds range of signed character.

#include <stdio.h>   int main() {     char = 60;     char j = 30;     char k = 10;     char l = (i*j)/k;     printf("%d ", l);      return 0; } 

the output of l 180 out of range char l, not getting error.

in other scenario if take same program instead of arithmetic function if put l=180 , try print wrong answer.

#include <stdio.h>   int main() {     char = 60;     char j = 30;     char k = 10;     char l = 180;     printf("%d ", l);      return 0; } 

the answer in 2nd case -76. can explain why? if virtually executing same thing getting different result.

are sure did , test described in first snippet? if run, overflows wraps around expected.

it makes big difference if did

char l = (i*j)/k; printf("%d ", l); 

and if did

printf("%d ", (i*j)/k); 

as in second case, compiler infer usage of int output, (the result not fit inside char without wrapping around obviously), , because intermediate computations made in int , not char.

in addition, consider result fits inside unsigned char, check case too.

in case, start thinking cases of problematic compiler too, after sure program executed should do.