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What does *int mean in Go? Detailed explanation of Go language pointer types
What does *int mean in Go? Detailed explanation of Go language pointer types
Apr 02, 2025 pm 01:06 PM
In-depth analysis of Go language pointer types: taking *int as an example
In Go language, *int represents a pointer to an integer variable. This is similar to the pointer concept in other languages, which stores a memory address, not the integer value itself. This article will explain in detail the usage and significance of Go language pointers in combination with sample code.
The following code snippet shows the application of Employee structure and new() function:
package main
import "fmt"
type Employee struct {
Id string
Name string
Age int
}
func main() {
e := Employee{"0", "P1", 33}
eNewed := new(Employee) // new() Returns a pointer to the Employee structure eNewed.Id = "1"
fmt.Printf("e: %T\n", e) // Output: e: main.Employee
fmt.Printf("eNewed: %T\n", eNewed) // Output: eNewed: *main.Employee
}
In the code, eNewed := new(Employee) uses the new() function to create a new Employee structure and returns its memory address, that is, a pointer to Employee structure. The output results show that the type of eNewed is *main.Employee . The * number is not a value operation here, but is used to declare a pointer type.
*int represents a pointer to data of type int ; similarly, *[]int represents a pointer to type []int (int slice). The * number is before the type name, indicating that this is a pointer type, and the memory unit it points to stores the value of the specified type. In *main.Employee , the * number indicates that eNewed variable stores the memory address of an Employee structure. Through this address, member variables in the Employee structure can be accessed and modified, such as eNewed.Id = "1" .
Understanding pointer types is crucial for Go programming, which improves program efficiency and flexibility, especially when dealing with large data structures and memory management.
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