Is this string addition a valid expression?

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I was curious if the compound assignment operators are valid for multiple parameters. My guess is += will have no side-effect but may not be the same case with "-=".

std::string a; 
a += "Hello" + " "+"World"; // doesn't compile
std::string name = "Old";
a += "Hello" + name +"World"; // compiles

This is not a valid expression because there is no operator + for string literals

"Hello" + " "+"World

(more precisely for pointers because in expressions string literals with rare exceptions are converted to pointers to their first symbols.)

Instead you could write

std::string a; 
( ( a += "Hello" ) += " " ) += "World";

But it would be more readable if to write

a += "Hello";
a += " ";
a += "World";

Or as @Bathsheba pointed out in a (valuable) comment to my answer you could use a user-defined string literal the following way

#include <string>
#include <iostream>

int main()
{
    using namespace std::string_literals;
    std::string a; 
    a += "Hello"s + " " + "World";

    std::cout << a << '\n';
}

As for this statement

a += "Hello" + name +"World";

then it can be rewriten using the operators defined for the class std::basic_string

template<class charT, class traits, class Allocator>
basic_string<charT, traits, Allocator>
operator+(const charT* lhs, const basic_string<charT, traits, Allocator>& rhs);

and

template<class charT, class traits, class Allocator>
basic_string<charT, traits, Allocator>
operator+(basic_string<charT, traits, Allocator>&& lhs, const charT* rhs);

like

a += operator +( operator +( "Hello", name ), "World" ); 

For example

#include <string>
#include <iostream>

int main()
{
    std::string a;
    std::string name = "Old";

    a += operator +( operator +( "Hello ", name ), " World" ); 

    std::cout << a << '\n';
}

Take into account that each operator returns an object of the type std::basic_string for which the operator + is defined. That is in each call of the operators there is present an object of the type std::basic_string as an argument.

Expressions and operators, It is thus the better alternative to provide defaults, when values like '' or 0 are valid values for the first expression, too. String operators. In addition to the comparison � Given a string str, the task is to check if the string is a valid identifier or not. In order to qualify as a valid identifier, the string must satisfy the following conditions: It must start with either underscore(_) or any of the characters from the ranges [‘a’, ‘z’] and [‘A’, ‘Z’]. There must not be any white space in the string.

The compound assignment operator += doesn't come into it.

You can use += with std::string because it defines the relevant overload, but it only gets one argument.

And you've tried to pass, as that argument, an expression constructed by +ing multiple string literals. This will not compile.

Some alternatives, in this toy case, which may inspire a solution for whatever is your real case:

Option 1
// Multiple calls to +=
std::string a;
a += "Hello";
a += " ";
a += "World";
Option 2
// String literal concatenation!
std::string a; 
a += "Hello" " " "World";
Option 3
// Repeated std::string concat
std::string a;
a += std::string("Hello") + " " + "World";
Option 4
// Same
using namespace std::string_literals;
std::string a; 
a += "Hello"s + " " + "World";
Option 5
// Don't bother
std::string a;
a += "Hello World";

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The expression a += "Hello" + " " + "World" is grouped as a += ("Hello" + " " + "World").

The right hand side is a set of 3 const char[] literals. These decay to const char* pointers when applied to binary addition. Since you can't add pointers, the compiler is required to issue a diagnostic.

(Note that the ostensibly equivalent expression a = a + "Hello" + " " + "World" is compilable.)

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2. Variables, expressions and statements — How to Think Like a , Not surprisingly, strings belong to the type str and integers belong to the type int. invalid syntax >>> class = "Computer Science 101" SyntaxError: invalid syntax Can you think of a property that addition and multiplication have that string� 1141 182 Add to List Share Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) + , - , or * between the digits so they evaluate to the target value.

Concatenation, In formal language theory and computer programming, string concatenation is the operation of In formal language theory and pattern matching (including regular expressions), the concatenation operation on strings is generalised to an � For information about the arithmetic + operator, see the Unary plus and minus operators and Addition operator + sections of the Arithmetic operators article. String concatenation When one or both operands are of type string , the + operator concatenates the string representations of its operands (the string representation of null is an empty

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Comments
  • Why didn't you just try to compile it to determine if it is valid?
  • @JesperJuhl To be fair, it's not because code compiles that it is valid.
  • @FrançoisAndrieux Of course not (as I know you know I know). But it would have let OP add the additional info: "it compiles" or "it doesn't compile with this error: ..." . And doing some basic testing yourself (like "does this even compile?") is a good habit to get into.
  • They (not me) probably thought you did not do enough research before asking the question. That is a bit subjective.
  • Welcome back Vlad!
  • @Bathsheba Thanks. I'm glad to meet you.:)
  • Or you could just write a += "Hello" " " "World" or even ``a += "Hello World"`, if you are feeling adventurous.
  • @DanM. You are right but the question is about the operator +.
  • @bruno: It has hidden depths. Note that if a += b really was replaced with a = a + b, then the code would compile.
  • Why does the second example compiles ?
  • @ash007 It's a language feature, albeit a little-known one.