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1. Error message
Defines a function that receives string parameters of type char*;
// 接收字符串参数并打印
void fun(char* str) {
cout << str << endl;
}
If you pass in a string constant, such as "Hello",
// 传入常量字符串
fun("Hello");
The complete code is as follows:
#include "iostream"
using namespace std;
// 接收字符串参数并打印
void fun(char* str) {
cout << str << endl;
}
int main() {
// 传入常量字符串
fun("Hello");
// 控制台暂停 , 按任意键继续向后执行
system("pause");
return 0;
};
Error message: This error is a compilation error;
Test.cpp(12,13): error C2664: "void fun(char *)": cannot convert parameter 1 from "const char [6]" to "char *"
Test.cpp(12,6): message: Conversion from string literal will lose const qualifier (see /Zc:strictStrings)
Test.cpp(5,6): message : see declaration of "fun"
Complete error report:
已启动生成…
1>------ 已启动生成: 项目: HelloWorld, 配置: Debug Win32 ------
1>Test.cpp
1>D:\002_Project\006_Visual_Studio\HelloWorld\HelloWorld\Test.cpp(12,13): error C2664: “void fun(char *)”: 无法将参数 1 从“const char [6]”转换为“char *”
1>D:\002_Project\006_Visual_Studio\HelloWorld\HelloWorld\Test.cpp(12,6): message : 从字符串文本转换将丢失 const 限定符(请参阅 /Zc:strictStrings)
1>D:\002_Project\006_Visual_Studio\HelloWorld\HelloWorld\Test.cpp(5,6): message : 参见“fun”的声明
1>已完成生成项目“HelloWorld.vcxproj”的操作 - 失败。
========== 生成: 成功 0 个，失败 1 个，最新 0 个，跳过 0 个 ==========
2. Problem analysis
This error only occurs in higher versions of Visual Studio, such as Visual Studio 2017, Visual Studio 2019 or higher;
In Visual Studio 2013, no error will be reported;
In the following fun function, a character array/string of type char* is received,
// 接收字符串参数并打印
void fun(char* str) {
cout << str << endl;
}
If the "Hello" parameter is passed in when calling, this is of const char* type, and the two parameter types do not match;
Ideas to solve the above problems:
- Modify function parameter type;
- Modify actual parameter type;
- Set the compatible configuration of the Visual Studio compilation environment;
3. Solution
1. Set the compatibility rules of Visual Studio
Set Visual Studio compatibility rules:
Right-click the solution in the Solution Explorer and select the last property option in the pop-up menu.
After opening, go to the configuration properties/C/C++/Language panel and check that the current compliance mode configuration is "Yes (/permissive-)",
Modify the configuration matching the pattern to "No",
At this point the program can be executed normally:
2. Modify the actual parameter type ①
If the function receives a string of type char*, then pass in the actual parameter of type char*, do not pass in a string of type const char*;
Convert the "Hello" string constant to the char* type, as shown in the following code example:
fun((char*)"Hello");
The complete code is:
#include "iostream"
using namespace std;
// 接收字符串参数并打印
void fun(char* str) {
cout << str << endl;
}
int main() {
fun((char*)"Hello");
// 控制台暂停 , 按任意键继续向后执行
system("pause");
return 0;
};
execution succeed :
3. Modify the actual parameter type ②
Put the string into a char array and pass the char array as an argument to the function;
char str[8] = "Hello";
fun(str);
Complete code example:
#include "iostream"
using namespace std;
// 接收字符串参数并打印
void fun(char* str) {
cout << str << endl;
}
int main() {
char str[8] = "Hello";
fun(str);
// 控制台暂停 , 按任意键继续向后执行
system("pause");
return 0;
};
execution succeed :
4. Modify the actual parameter type ③
Previously, forced type conversion in C language was used. Here, forced type conversion in C++ is used to convert constants into non-constants, and the const_cast operator is used for conversion.
fun(const_cast<char*>("Hello"));
Complete code example:
#include "iostream"
using namespace std;
// 接收字符串参数并打印
void fun(char* str) {
cout << str << endl;
}
int main() {
fun(const_cast<char*>("Hello"));
// 控制台暂停 , 按任意键继续向后执行
system("pause");
return 0;
};
execution succeed :
5. Modify formal parameter type
This problem can also be solved by changing the formal parameter of type char* in the function to type const char*;
Complete code example:
#include "iostream"
using namespace std;
// 接收字符串参数并打印
void fun(const char* str) {
cout << str << endl;
}
int main() {
fun("Hello");
// 控制台暂停 , 按任意键继续向后执行
system("pause");
return 0;
};
Results of the :
