[Error] ambiguous overload for 'operator=' (operand types are 'std::string {aka std::basic_string<char>}' and 'std::basic_istream<char>')
时间: 2023-10-15 11:02:11 浏览: 212
这个错误是因为你试图将一个 `std::basic_istream<char>` 类型的对象赋值给一个 `std::string` 类型的对象,而 `std::string` 类型的赋值运算符并不支持这种操作。
可能是因为你在尝试从输入流中读取数据,并将其赋值给一个字符串对象,但是你使用了错误的赋值语法。正确的做法是使用输入流的读取函数来读取数据,然后将其存储到一个字符串对象中。
例如,下面的代码可以从标准输入流中读取一行文本,并将其存储到一个字符串对象中:
```c++
#include <iostream>
#include <string>
int main() {
std::string input;
std::getline(std::cin, input);
std::cout << "Input: " << input << std::endl;
return 0;
}
```
在这个例子中,`std::getline()` 函数从 `std::cin` 输入流中读取一行文本,并将其存储到 `input` 字符串对象中。这个函数会自动处理字符串的长度和换行符等细节,因此你不需要担心这些问题。
相关问题
ambiguous overload for 'operator=' (operand types are 'std::string {aka std::basic_string<char>}' and 'long long int')
This error message occurs when you try to assign a long long integer value to a std::string variable using the assignment operator (=). Since there is no overloaded operator= function that takes both a std::string and a long long integer argument, the compiler cannot determine which one to use.
To fix this error, you need to convert the long long integer to a std::string before assigning it to the variable. You can use std::to_string() function to convert the long long integer to a string as shown below:
```
std::string str = std::to_string(100LL); // convert long long integer to string
```
Alternatively, you can use a constructor of std::string that takes a const char* argument to create a string from a C-style string representation of the long long integer:
```
std::string str = std::string("100"); // create string from C-style string
```
In either case, the resulting string can be assigned to the std::string variable without any ambiguity.
error: use of overloaded operator '<<' is ambiguous (with operand types 'std::stringstream' (aka 'basic_stringstream<char>') and 'const std::basic_string<char>::value_type *' (aka 'const char *')) stream << id.toString().c_str();
这个错误的原因是在使用stringstream的时候,对于const char*类型的字符串,编译器无法确定应该使用哪个重载的<<运算符。解决方法是将const char*类型的字符串转换为std::string类型再进行输出,或者使用static_cast强制类型转换。例如:
```
stream << static_cast<std::string>(id.toString());
```
或者
```
stream << std::string(id.toString());
```
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