redeclaration of"i"with no linkage
时间: 2024-05-24 09:12:44 浏览: 240
As an AI language model, I can explain that a redeclaration of "i" with no linkage refers to creating a new variable or object called "i" within a specific scope that has no connection or access to any other "i" variables or objects outside of that scope. This means that any changes or modifications made to this specific "i" will not affect any other "i" variables or objects in other scopes.
相关问题
redeclaration of "rawtime" with no linkage
This error message typically occurs when you try to declare a variable with the same name multiple times in the same scope without any external linkage. In C, each variable must have a unique name within its scope.
To resolve this issue, you should ensure that you only declare the variable "rawtime" once in the current scope. If you need to use the variable in different parts of your code, you can either declare it in separate scopes or use different names for each declaration.
Here's an example to illustrate this:
```c
#include <time.h>
int main() {
time_t rawtime; // Declare "rawtime" once
// Rest of your code...
{
time_t rawtime; // Another declaration of "rawtime" in a different scope
// Rest of your code...
}
return 0;
}
```
By separating the declarations into different scopes, you can avoid redeclaring the variable without linkage error.
[Error] redeclaration of 'i' with no linkage
在C/C++编程中,"redeclaration of 'i' with no linkage" 错误通常意味着你在程序的不同作用域中多次声明了同名的变量 'i',但是没有提供外部链接(extern)。在一个作用域内,全局变量、静态局部变量以及函数内的局部变量都各自拥有独立的作用域,如果在同一作用域内再次声明一个已经存在的未链接变量,就会导致这样的错误。
例如:
```c
int i;
void func() {
int i; // 在这里,我们又声明了一个局部变量 i,但没有 extern 关键字表示它是全局的
// ...
}
// 或者
int main() {
int i;
static int i; // 在main函数内部,静态局部变量 i 和之前的全局变量 i 也构成了重复声明
// ...
}
```
要修复此错误,你需要确保每个变量在整个程序范围内只声明一次,并且如果需要在不同作用域间共享变量,应该使用 extern 关键字声明全局变量。例如:
```c
int global_i; // 正确的全局变量声明
void func() {
extern int global_i; // 明确告知这是一个外部引用
int local_i; // 现在local_i是一个新的局部变量
// ...
}
int main() {
global_i = 10; // 可以在main中访问全局变量
// ...
}
```
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