Bitwise OR not working C [duplicate]

This question already has answers here: OR operator in C not working (4 answers) Closed 9 years ago.

Please help me with this code. I don't know why the output is not 8.

#include <stdio.h>
#include <stdlib.h>
int main(){
    char c;
    c = c & 0;
    printf("The value of c is %d", (int)c);
    int j = 255;
    c = (c | j);
    int i =0;
    while( (c &a开发者_开发百科mp; 1) == 1){
        i++;
        c = c>>1;
    }
    printf("the value of i is %d",i);
    system("pause");
    return 0;
}

To all the people who gave me -1, I tried printing the values. But it goes into an infinite loop.

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As you learnt from your previous question, char is signed on your platform. Therefore, at the beginning of your while loop, c has a negative value.

Right-shifting a negative value is implementation-defined. On your platform, I imagine it is doing an arithmetic shift.

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This code has undefined behavior on the very first statement. c is uninitialized and has an indeterminate value, and using an object with indeterminate value, even to bitwise-and or multiply it by zero, results in undefined behavior.

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Try using an unsigned char. The standard C char can be signed (and quite often is on PC).