string-concatenation

I'm running this in node.js: > x = { 'foo' : 'bar' } { foo: 'bar' } > console.log(x) { foo: 'bar' } undefined > console.log("hmm: " + x) hmm: [object Object] undefined What I don't understand is why console.log(x) "pretty-prints" the object, whereas string concatenation "ugly-prints" it. And more importantly, what's the best way to make it print hmm: { foo: 'bar' } ? The + x coerces the object x into a string, which is just [object Object] : http://jsfiddle.net/Ze32g/ The pretty printing is a very nice and probably very complex underlying code that someone implemented as part of the console

I know it is a common issue, but looking for references and other material I don't find a clear answer to this question. Consider the following code: #include <string> // ... // in a method std::string a = "Hello "; std::string b = "World"; std::string c = a + b; The compiler tells me it cannot find an overloaded operator for char[dim] . Does it mean that in the string there is not a + operator? But in several examples there is a situation like this one. If this is not the correct way to concat more strings, what is the best way? Your code, as written, works. You’re probably trying to achieve

I need to concatenate two const chars like these: const char *one = "Hello "; const char *two = "World"; How might I go about doing that? I am passed these char* s from a third-party library with a C interface so I can't simply use std::string instead. In your example one and two are char pointers, pointing to char constants. You cannot change the char constants pointed to by these pointers. So anything like: strcat(one,two); // append string two to string one. will not work. Instead you should have a separate variable(char array) to hold the result. Something like this: char result[100]; //

This question already has an answer here: StringBuilder vs String concatenation in toString() in Java 19 answers What is the benefit and trade-off of using a string builder over pure string concatenation? new StringBuilder(32).append(str1) .append(" test: ") .append(val) .append(" is changed") .toString(); vs say str1 + " test: " + val + " is changed". str1 is a random 10 character string. str2 is a random 8 character string. In your particular example, none because the compiler internally uses StringBuilder s to do String concatenation. If the concatenation occurred in a loop, however, the

This question already has an answer here: String concatenation: concat() vs “+” operator 11 answers For string concatenation we can use either the concat() or concat operator (+) . I have tried the following performance test and found concat() is faster and a memory efficient way for string concatenation. String concatenation comparison for 100,000 times : String str = null; //------------Using Concatenation operator------------- long time1 = System.currentTimeMillis(); long freeMemory1 = Runtime.getRuntime().freeMemory(); for(int i=0; i<100000; i++){ str = "Hi"; str = str+" Bye"; } long time2