const char*连接

Question

我需要连接两个这样的const字符:

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

我该怎么做呢？

我char*从带有C接口的第三方库中传递了这些,所以我不能简单地使用它std::string.

Answer 1

在您的示例中,一个和两个是char指针,指向char常量.您无法更改这些指针指向的char常量.所以像:

strcat(one,two); // append string two to string one.

不管用.相反,你应该有一个单独的变量(char数组)来保存结果.像这样的东西:

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.

Answer 2

C方式:

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

C++方式:

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

编译时方式:

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

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

Answer 3

如果您使用的是C++,为什么不使用std::stringC风格的字符串呢？

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

std::string three= one+two;

如果需要将此字符串传递给C函数,只需传递即可 three.c_str()

Answer 4

使用std::string:

#include <string>

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

Answer 5

更新:改 string total = string(one) + string(two); 到string total( string(one) + two );性能方面的原因(避免串两个临时串共建设)

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()

Answer 6

还有一个例子:

// 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;

但除非您特别不想或不能使用C++标准库,否则使用std::string可能更安全.

Answer 7

首先,您必须创建一些动态内存空间.然后你可以将两个字符串strcat到它中.或者您可以使用c ++"string"类.老派的C方式:

  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);

新的C++方式

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

Answer 8

看起来你正在使用带有C库的C++,因此你需要使用它const char *.

我建议把它们包装const char *成std::string:

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

Answer 9

如果您不知道字符串的大小，则可以执行以下操作：

#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;
}