Home/ Questions/Q 6948151
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-27T13:47:05+00:00

The following is excerpted from Writing solid code , page 115. int strcmp( const

  • 0

The following is excerpted from Writing solid code, page 115.

int strcmp( const char *strLeft, const char *strRight )
{
  for( NULL; *strLeft == *strRight; strLeft ++ ,strRight ++ )
      if( strLeft == ‘\0’ ) 
          return(0);
  return ( (*strLeft<*strRight)?-1:1 );
}

The moment you use <, or any other operator that use sign information,
you force the compiler to generate nonprtable code

What does it(Bold line) mean? I know that

  1. right shifting sign integer is non-portable.
  2. comparing sign integer and unsigned integer is non-portable

Why is comparing two sign integer non-portable?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    The ‘char’ type in C can be signed or unsigned. So here, if the strings are {0x81, 0} and {0x32, 0}, then if chars are signed 0x81 will be interpreted as a negative number (and the result will be that the first string compares less), and if chars are unsigned it’ll be interpreted as a positive number (so the second string will compare less). This is non-portable in the sense that the results differ based on the compiler you’ve used.

    • 0