const char* concatenation

问题

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.

回答1:

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];   // array to hold the result.

strcpy(result,one); // copy string one into the result.
strcat(result,two); // append string two to the result.

回答2:

The C way:

char buf[100];
strcpy(buf, one);
strcat(buf, two);

The C++ way:

std::string buf(one);
buf.append(two);

The compile-time way:

#define one "hello "
#define two "world"
#define concat(first, second) first second

const char* buf = concat(one, two);

回答3:

If you are using C++, why don't you use std::string instead of C-style strings?

std::string one="Hello";
std::string two="World";

std::string three= one+two;

If you need to pass this string to a C-function, simply pass three.c_str()

回答4:

Using std::string:

#include <string>

std::string result = std::string(one) + std::string(two);

回答5:

Update: changed string total = string(one) + string(two); to string total( string(one) + two ); for performance reasons (avoids construction of string two and temporary string total)

const char *one = "Hello ";
const char *two = "World";

string total( string(one) + two );    // OR: string total = string(one) + string(two);
// string total(move(move(string(one)) + two));  // even faster?

// to use the concatenation in const char* use
total.c_str()

回答6:

One more example:

// calculate the required buffer size (also accounting for the null terminator):
int bufferSize = strlen(one) + strlen(two) + 1;

// allocate enough memory for the concatenated string:
char* concatString = new char[ bufferSize ];

// copy strings one and two over to the new buffer:
strcpy( concatString, one );
strcat( concatString, two );

...

// delete buffer:
delete[] concatString;

But unless you specifically don't want or can't use the C++ standard library, using std::string is probably safer.

回答7:

It seems like you're using C++ with a C library and therefore you need to work with const char *.

I suggest wrapping those const char * into std::string:

const char *a = "hello "; 
const char *b = "world"; 
std::string c = a; 
std::string d = b; 
cout << c + d;

回答8:

First of all, you have to create some dynamic memory space. Then you can just strcat the two strings into it. Or you can use the c++ "string" class. The old-school C way:

  char* catString = malloc(strlen(one)+strlen(two)+1);
  strcpy(catString, one);
  strcat(catString, two);
  // use the string then delete it when you're done.
  free(catString);

New C++ way

  std::string three(one);
  three += two;

回答9:

If you don't know the size of the strings, you can do something like this:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(){
    const char* q1 = "First String";
    const char* q2 = " Second String";

    char * qq = (char*) malloc((strlen(q1)+ strlen(q2))*sizeof(char));
    strcpy(qq,q1);
    strcat(qq,q2);

    printf("%s\n",qq);

    return 0;
}

回答10:

You can use strstream. It's formally deprecated, but it's still a great tool if you need to work with C strings, i think.

char result[100]; // max size 100
std::ostrstream s(result, sizeof result - 1);

s << one << two << std::ends;
result[99] = '\0';

This will write one and then two into the stream, and append a terminating \0 using std::ends. In case both strings could end up writing exactly 99 characters - so no space would be left writing \0 - we write one manually at the last position.

回答11:

const char* one = "one";
const char* two = "two";
char result[40];
sprintf(result, "%s%s", one, two);

回答12:

Connecting two constant char pointer without using strcpy command in the dynamic allocation of memory:

const char* one = "Hello ";
const char* two = "World!";

char* three = new char[strlen(one) + strlen(two) + 1] {'\0'};

strcat_s(three, strlen(one) + 1, one);
strcat_s(three, strlen(one) + strlen(two) + 1, two);

cout << three << endl;

delete[] three;
three = nullptr;

来源：https://stackoverflow.com/questions/1995053/const-char-concatenation

标签

c++

string-concatenation