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up vote 16 down vote favorite 
std::string tmp;
tmp +=0;//compile error:ambiguous overload for 'operator+=' (operand types are 'std::__cxx11::string {aka std::__cxx11::basic_string<char>}' and 'int')
tmp +=1;//ok
tmp += '\0';//ok...expected
tmp +=INT_MAX;//ok
tmp +=int(INT_MAX);//still ok...what?

The first one argues that passing integer as argument, right? Why others passes compilation?I tested on Visual C++ and g++, and I got the same result above. So I believe I miss something defined by standard. What is it?

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3  
int will be converted to char. but for 0 its difficult to say whether its NULL or int 0 – Praveen 10 hours ago
up vote 23 down vote accepted

The problem is that a literal 0 is a null pointer constant. The compiler doesn't know if you meant:

std::string::operator +=(const char*);  // tmp += "abc";

or

std::string::operator +=(char);         // tmp += 'a';

(better compilers list the options).

The workround (as you have discovered) is to write the append as:

tmp += '\0';

(I assume you didn't want the string version - tmp += nullptr; would be UB at runtime.)

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Should the compiler really consider +=(char)? – Bathsheba 10 hours ago
4  
Yes. char is a integral type, and an integer is implicitly convertable to char. +=(char) is the overload that is being invoked in all the non-ambiguous cases. – Martin Bonner 10 hours ago
up vote 5 down vote

The 0 literal is implicitly convertible to all pointer types (resulting in their respective null pointer constants). Therefore it results in two equally valid conversion sequences for matching std::strings appending operator.

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