A classic interview question in C language under Linux

Question

If you use the GCC compiler to execute the following program under Linux, what is the output result?

#include<stdio.h>
int main(){
    char c=127;
    printf("%d",++c);
    printf("%d",++c);
    return 0;
}

Just know that it involves type conversion, data truncation and filling. But don’t know the specific explanation?

Original question source: Several classic interview questions in C language under Linux

Answer 1

The length of

char is 1 byte, and most machines treat it as a signed number, so its representation range is [-128, 127] (see "In-depth Understanding of Computer Systems" P27 ~P28). So, when you assign 127 to c, you execute ++c, which causes an overflow because it only has one byte.

represents 127 in the machine, and it turns into binary like this 01111111. You can see that when you add 1, the result becomes 10000000. Since within the computer, negative numbers are represented by two’s complement, So it becomes -128. Then ++c, it’s -127.

As for the different types, they all behave the same inside the computer, which is a piece of memory. So type is not a limitation.

Answer 2

This question tests the compiler, not the language.
Define 3 variables:

char c; // 表示一个 ascii字符
signed char s_c;   //有符号 1字节 整数
unsigned char u_c; //无符号 1字节 整数

In C language, when c participates in calculation, whether c is converted into s_c or u_c is decided by the compiler.

gcc considers c to be signed, and subsequent calculations and outputs are processed as signed numbers.