Golang的if语句无法识别int32类型的0。

huangapple
go评论121阅读模式
英文:

Golang if statement not seeing 0 for int32

问题

我有一个用golang编写的proto3/grpc函数。在一个switch语句中，if语句无法将int32的值0视为0。我在之前打印了该值，确实是0，但是if语句仍然执行。在我的代码中，我在注释中给出了输出。我知道对于int类型，nil值是0。如果我为lname和fname设置一个值，它们会按预期工作。非常感谢任何帮助。以下是我的输出：

  1. map[fname: lname: email: id:0]
  2. 0
  3. id = $1

以下是我的代码：

  1. func (s *server) GetUsers(ctx context.Context, in *userspb.User) (*userspb.Users, error) {
  2.     flds := make(map[string]interface{})
  3.     flds["id"] = in.Id // 0
  4.     flds["fname"] = in.Fname // "" (空)
  5.     flds["lname"] = in.Lname // "" (空)
  6.     flds["email"] = in.Email // "" (空)
  8.     fmt.Println(flds) //map[lname: email: id:0 fname:]
  10.     var where bytes.Buffer
  12.     n := 0
  13.     for _, v := range flds {
  14.         switch v.(type) {
  15.         case string:
  16.             if v != "" {
  17.                 n++
  18.             }
  19.         case int, int32, int64:
  20.             if v != 0 {
  21.                 n++
  22.             }
  23.         }
  24.     }
  26.     calledvariables := make([]interface{}, 0)
  28.     i := 1
  29.     for k, v := range flds {
  31.         switch v.(type) {
  32.         case string:
  33.             if v != "" {
  34.                 if i != 1 {
  35.                     where.WriteString(" AND ")
  36.                 }
  37.                 ist := strconv.Itoa(i)
  38.                 where.WriteString(k + " = $" + ist)
  39.                 calledvariables = append(calledvariables, v)
  40.                 i++
  41.             }
  42.         case int, int32, int64, uint32, uint64:
  43.             /////// 这个if语句有问题（v打印的值是0，但它在if语句中）
  44.             if v != 0 {
  45.                 fmt.Println(v) // 0
  46.                 if i != 1 {
  47.                     where.WriteString(" AND ")
  48.                 }
  49.                 ist := strconv.Itoa(i)
  50.                 where.WriteString(k + " = $" + ist)
  51.                 calledvariables = append(calledvariables, v)
  52.                 i++
  53.             }
  54.         }
  55.     }
  57.     fmt.Println(where.String()) // id = $1
  58.     ...
英文:

I have proto3/grpc function written in golang. There is a if statement written in a switch that does't see an int32 as a value of 0 when the value is a 0. I print the value before and it's a 0 but the if statement runs anyway. In my code below I have the output in a comment. I know for a int the nil value is a 0. If I place a value for the lname, fname they work as it should. Any help appreciated. Here is my output:

  1. map[fname: lname: email: id:0]
  2. 0
  3. id = $1

Here is my code:

  1. func (s *server) GetUsers(ctx context.Context, in *userspb.User) (*userspb.Users, error) {
  2. 	flds := make(map[string]interface{})
  3. 	flds["id"] = in.Id // 0
  4. 	flds["fname"] = in.Fname // "" (empty)
  5. 	flds["lname"] = in.Lname // "" (empty)
  6. 	flds["email"] = in.Email // "" (empty)
  8. 	fmt.Println(flds) //map[lname: email: id:0 fname:]
  10. 	var where bytes.Buffer
  12. 	n := 0
  13. 	for _, v := range flds {
  14. 		switch v.(type) {
  15. 		case string:
  16. 			if v != "" {
  17. 				n++
  18. 			}
  19. 		case int, int32, int64:
  20. 			if v != 0 {
  21. 				n++
  22. 			}
  23. 		}
  24. 	}
  26. 	calledvariables := make([]interface{}, 0)
  28. 	i := 1
  29. 	for k, v := range flds {
  31. 		switch v.(type) {
  32. 		case string:
  33. 			if v != "" {
  34. 				if i != 1 {
  35. 					where.WriteString(" AND ")
  36. 				}
  37. 				ist := strconv.Itoa(i)
  38. 				where.WriteString(k + " = $" + ist)
  39. 				calledvariables = append(calledvariables, v)
  40. 				i++
  41. 			}
  42. 		case int, int32, int64, uint32, uint64:
  43. 			/////// THIS IF STATMENT IS THE ISSUE the ( v is printing the value of 0 and it's in the if statement )
  44. 			if v != 0 {
  45. 				fmt.Println(v) // 0
  46. 				if i != 1 {
  47. 					where.WriteString(" AND ")
  48. 				}
  49. 				ist := strconv.Itoa(i)
  50. 				where.WriteString(k + " = $" + ist)
  51. 				calledvariables = append(calledvariables, v)
  52. 				i++
  53. 			}
  54. 		}
  55. 	}
  57. 	fmt.Println(where.String()) // id = $1
  58.     ...

答案1

得分: 5

因为字面量0不是相同的类型。如果你这样写：

  1. if v != int32(0) {

当值是int32类型时，它会按预期工作。不幸的是，你将所有的整数类型组合在一个case中，这将使得正确处理这个问题变得困难/不便。你可以尝试使用反射，在运行时将值与其类型的零值进行比较，使用reflect.Zero。

英文:

Because the literal 0 is not the same type. If you do:

  1. if v != int32(0) {

When the value is an int32, it works as expected. Unfortunately, you're combining all the int types in a single case, which will make this difficult/unwieldy to handle correctly. You could probably work something out using reflection to compare the value against the zero value for its type at runtime, using reflect.Zero.

huangapple
  • 本文由 发表于 2017年5月30日 21:16:33
  • 转载请务必保留本文链接：https://go.coder-hub.com/44263381.html
  • go
  • grpc
  • protocol-buffers
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定