为什么在上班时不检查零？

Question

In the first code example, I get errors for the "if pr != nil" line:

for sup, _ := range supervisorToColor {
        pr := emailToPerson[sup]
        // The line below causes the compilation error:
        // ./myprog.go:1046: missing condition in if statement
        // ./myprog.go:1046: pr != nil evaluated but not used
        if pr != nil 
        {   
          local := peopleOnFloor[pr.Email]
          sp := &Super{pr, local}
          names = append(names, sp) 
        }   
}   

If I comment out the nil check if statement, it compiles fine:

for sup, _ := range supervisorToColor {
        pr := emailToPerson[sup]
        // if pr != nil 
        // {
          local := peopleOnFloor[pr.Email]
          sp := &Super{pr, local}
          names = append(names, sp) 
        // }
}   

At first I was inclined to think it was some syntax error earlier in the code, but the fact that it works when I comment out the lines makes me think it's something else.

emailToPerson is of type map[string]*Person where Person is a struct

Thanks in advance. Apologies if this turns out to be something incredibly simple.

Answer 1

The open curly brace needs to be on the same line as the if:

if pr != nil { 

From the Go spec on semicolons:

The formal grammar uses semicolons ";" as terminators in a number of productions. Go programs may omit most of these semicolons using the following two rules:

1. When the input is broken into tokens, a semicolon is automatically inserted into the token stream immediately after a line's final token if that token is

   • an identifier
   • an integer, floating-point, imaginary, rune, or string literal
   • one of the keywords break, continue, fallthrough, or return
   • one of the operators and delimiters ++, --, ), ], or }

2. To allow complex statements to occupy a single line, a semicolon may be omitted before a closing ")" or "}".

This means that your code was equivalent to:

if pr != nil;
{
    // ...
}