[Solved] error: invalid operands of types ‘const char [6]‘ and ‘const char [6]‘ to binary ‘operator+‘



When using strings in C++, we habitually use + to connect two strings enclosed in "". The error is reported: error: invalid operators of types' const char [6] 'and' const char [6] 'to binary' operator+',

1. Scenarios

	//std::string str = "hello" + "world"; // error: invalid operands of types 'const char [6]' and 'const char [6]' to binary 'operator+'
	//std::string str = std::string("hello") + " world"; // correct
	//std::string str = "hello" + std::string("world"); //correct
	std::string str = "hello"" world"; //correct

    cout << "str=" << str << endl;

When using the+operator to connect two strings wrapped with "", an error occurs. The reason is that in C++, the strings enclosed by "" are regarded as const char* types, rather than string types.

2. Solution

Refer to the explanation on stackoverflow. For details, please read error: invalid operators of types’ const char * ‘and’ const char * ‘to binary’ operator+’

1. Explicitly declare one of them as std::string type (recommended)

std::string str = std::string("hello") + " world";

2. Remove the+and let the compiler splice strings (not recommended)

std::string str = "hello"" world";

Most compilers automatically splice strings enclosed by "", but it is not guaranteed that all compilers are normal. It is recommended to display strings declared as string types and then perform the+operation.

3. Validation

	std::string str1 = std::string("hello") + " world" + "!"; // correct
    std::string str2 = "hello" + std::string(" world") + "!"; //correct
    std::string str3 = "hello"" world" "!"; //correct

    cout << "str1=" << str1 << endl;
    cout << "str2=" << str2 << endl;
    cout << "str3=" << str3 << endl;

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