What is it mean by exit code 139 in C programming when i compile a code? [duplicate]

here error code 139 indicates segment fault that is arising because you haven't used format specifiers in the code the correct code will be

#include <stdio.h> #include <string.h> int main(void) { char ch = 'A'; short s = ch; printf("%d",s); printf("\n________________________\n"); short s2 = 67; char ch2 = (char)s2; printf("%c",ch2); return 0; } 

Your problem is the line

printf(s); 

printf expects its first argument to be a format string, and the type of the argument must be const char *. It's trying to treat s as the address of a string, but the value of 'A' (65 in ASCII) is not a valid address, hence the error.

To print the value of s, you need to do something like

printf( "%hd", s ); 

The format specifier %hd tells scanf to expect a short argument, and to print the string equivalent of its value as decimal digits ("65").

You have a similar issue in the second printf call.

https://www.cplusplus.com/reference/cstdio/printf/ expects a pointer to 0-terminated string. You are passing some short value (A, wich is 65) that is interpreted as an address.

You can use type casting and format specifiers as follows:

#include <stdio.h> #include <string.h> int main(void) { char ch = 'A'; short s = ch; //short format specifier printf("%hd",s); printf("\n________________________\n"); short s2 = 67; //int to char char ch2 = (char)s2; //char format specifier printf("%c",ch2); return 0; }