In Go language, why can math.Sqrt(2) run normally while math.Sqrt(r) report an error?

Detailed explanation of Go language math.Sqrt function parameter types and common errors

This article uses an example to explain the parameter type problem of the math.Sqrt function in Go language, and why math.Sqrt(2) can run normally while math.Sqrt(r) will report an error.

Let's look at a piece of Go code:

 package main

import (
    "fmt"
    "math"
)

func main() {
    r := 2
    f := 2.0

    fmt.Printf("%T %v\n", r, r) // int 2
    fmt.Printf("%T %v\n", f, f) // float64 2

    fmt.Printf("%T %v\n", math.Sqrt(2), math.Sqrt(2)) // float64 1.4142135623730951
    fmt.Printf("%T %v\n", math.Sqrt(r), math.Sqrt(r)) // compile error
    fmt.Printf("%T %v\n", math.Sqrt(f), math.Sqrt(f)) // float64 1.4142135623730951
}

In the code, r is an integer type and f is a floating point type. math.Sqrt(2) can run normally because 2 is a typeless literal in Go language. The compiler will implicitly convert it to float64 type according to the context, meeting the requirements of the math.Sqrt function.

However, math.Sqrt(r) will report an error because r is a variable declared as int and cannot be directly converted to float64 type. The math.Sqrt function expects a float64 parameter, while r is int type, resulting in type mismatch.

math.Sqrt(f) can run normally because f itself is float64 type.

Summarize:

The typeless literal feature of Go allows the compiler to make type inference based on the context. However, variables of declared types must strictly match the function parameter type. To avoid errors like math.Sqrt(r) , r should be explicitly converted to float64 type:

 fmt.Printf("%T %v\n", math.Sqrt(float64(r)), math.Sqrt(float64(r))) // float64 1.4142135623730951

Understanding the Go language's type system and the nature of typeless literals is crucial for writing high-quality, error-free Go code. Remember, when calling a function, always make sure that the parameter type matches the type declared by the function.

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