Why won't these strings concatenate in C++?

Question

I have an example of two test programs I wrote in C++. The first one works fine, the first one errors. Please help me to explain what is going on here.

#include <iostream>
#include <string>
#include <stdint.h>
#include <stdlib.h>
#include <fstream>
using namespace std;

string randomStrGen(int length) {
static string charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
string result;
result.resize(length);
for (int32_t i = 0; i < length; i++)
    result[i] = charset[rand() % charset.length()];
return result;
}

int main()
{
ofstream pConf;
pConf.open("test.txt");
pConf << "rpcuser=user\nrpcpassword=" 
     + randomStrGen(15)
     + "\nrpcport=14632"
     + "\nrpcallowip=127.0.0.1"
     + "\nport=14631"
     + "\ndaemon=1"
     + "\nserver=1"
     + "\naddnode=107.170.59.196";
pConf.close();
return 0;
}

It opens 'test.txt' and writes the data, no problem. This, however, does not:

#include <iostream>
#include <string>
#include <stdint.h>
#include <stdlib.h>
#include <fstream>
using namespace std;

string randomStrGen(int length) {
static string charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
string result;
result.resize(length);
for (int32_t i = 0; i < length; i++)
    result[i] = charset[rand() % charset.length()];
return result;
}

int main()
{
ofstream pConf;
pConf.open("test.txt");
pConf << "rpcuser=user\n"
     + "rpcpassword=" 
     + randomStrGen(15)
     + "\nrpcport=14632"
     + "\nrpcallowip=127.0.0.1"
     + "\nport=14631"
     + "\ndaemon=1"
     + "\nserver=1"
     + "\naddnode=107.170.59.196";
pConf.close();
return 0;
}

The only difference in the second program is that 'rpcpassword' has been moved to the next line.

matthew@matthew-Satellite-P845:~/Desktop$ g++ test.cpp 
test.cpp: In function ‘int main()’:
test.cpp:23:6: error: invalid operands of types ‘const char [14]’ and ‘const char [13]’ to binary ‘operator+’ 
  + "rpcpassword="

The difference is std::string has operator+ overloaded for concatenation with const char*. — user3920237
– user3920237, Commented Feb 6, 2015 at 1:14
Aasmund's answer's right, but incidentally adjacent string literals gets concatenated by the compiler in a pretty early phase (e.g. "abc" "def" becomes "abcdef", even if they're on different lines in the source). So, if you remove all the '+'s except those before and after randomStrGen(15), it will not only compile properly but use less memory and run faster. — Tony Delroy
– Tony Delroy, Commented Feb 6, 2015 at 1:21

Aasmund Eldhuset · Accepted Answer · 2015-02-06 01:15:40Z

5

A string literal ("foo") in C++ is not of the type string; it is of the type const char[x], where x is the length of the string literal plus 1. And character arrays can't be joined with +. However, a character array can be joined with a string, and the result is a string, which may further be joined with character arrays. Therefore, "a" + functionThatReturnsString() + "b" works, but "a" + "b" does not. (Keep in mind that + is left associative; it is applied to the two leftmost operands first, and then to the result and the third operand, and so on.)

answered Feb 6, 2015 at 1:15

Aasmund Eldhuset

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4 Comments

user3920237 Over a year ago

The OP should just remove the + as two consecutive string literals are automatically concatenated.

user3920237 Over a year ago

@user2676680 There are many non-member overloads for operator+ which take a const char* and a std::string. However, there is no built-in operator+ which takes two const char*s.

Praetorian Over a year ago

@user2676680 The last 2 sentences explain why the first one works

5gon12eder Over a year ago

In case this did not become clear from @remyabel's comment: They are concatenated during translation of the program so it also has zero run-time overhead. If you don't have literals, you can replace your + with << but it will be less efficient in your current case.

Boyd Stephen Smith Jr. · Accepted Answer · 2015-02-06 01:20:27Z

4

"rpcuser=user\nrpcpassword=" + randomStrGen(15) + "\nrpcport=14632" groups like ("rpcuser=user\nrpcpassword=" + randomStrGen(15)) + "\nrpcport=14632". Here, + is always used with an argument of a class type, so you get std::string::operator+ after overload resolution.

"rpcuser=user\n" + "rpcpassword=" + randomStrGen(15) groups like ("rpcuser=user\n" + "rpcpassword=") + randomStrGen(15). In this case, the first + is used on two non-class types, so it's not overloaded and the language doesn't define + for two const char [] values. (I come from old C, so I was a bit it didn't just add them as char *s and give you a nice SIGSEGV at runtime.)

answered Feb 6, 2015 at 1:20

Boyd Stephen Smith Jr.

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2 Comments

Matthew Darnell Over a year ago

Both answers were very helpful, your groupings with the parentheses made it click. I don't like the fact that it is overloading without me knowing about it, it seems this will lead to programmers not knowing exactly what is going on. I will make sure to change my use of string literals.

Aasmund Eldhuset Over a year ago

I should have used parentheses to make my point more clear, so this gets an upvote from me too :-) (As long as you're using C++, you better get used to overloading...)

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