Arithmectic operator in c,modulus(%) giving warning. why? [duplicate]

Question

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How do I print the percent sign(%) in C? [duplicate] (4 answers)

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int main(void)
{
    int a=12,b=3;
    printf("\n a+b = %i\n",a+b);
    printf("\n a-b = %i\n",a-b);
    printf("\n a*b = %i\n",a*b);
    printf("\n a/b = %i\n",a/b);
    printf("\n a%b = %i\n",a%b);//conversion type error 
}

The modulus part is giving the warning as Unknown conversion type character 'b' in format.

Answer 1

It is printf giving the warning, scape the modulus character with another modulus:

printf("\n a%%b = %i\n",a%b);

As you may see in the manual: printf(3) there's no b flag character, so when printf find your %b in your string it doesn't know what to do. Since you don't want any formatting in that case, just include the % character in your string, you just need to scape the % character with another % character as in the example above.