C字符串拼接

Question

One of the answers到Why do you not use C for your web apps?包含以下內容：

For the C crap example below:

const char* foo = "foo"; 
const char* bar = "bar"; 
char* foobar = (char*)malloc(strlen(foo)+strlen(bar)+1); 
strcpy(foobar, foo); 
strcat(foobar, foo); 

Actually, constants CAN AND SHOULD be concatenated naturally in C:

const char foo[] = "foo"; 
const char bar[] = "bar"; 
char foobar[] = foo bar; // look Ma, I did it without any operator! 

And using [] instead of * will even let you modify the string, or find their length:

int foo_str_len = sizeof(foobar)-1; 

So, PLEASE, before you (falsely) claim that C is difficult to use with strings, learn how to use C.

---

我已經嘗試過自己，但得到一個錯誤：

expected ‘,’ or ‘;’ before string constant

所以我的問題是：做我需要告訴編譯器，爲了使這個工作或上述帖子簡直是錯誤的？請注意，我知道用C連接字符數組的其他方法。

Answer 1

(char*)malloc

決不強制轉換的C.閱讀this和this的malloc的結果。

Actually, constants CAN AND SHOULD be concatenated naturally in C

否，字符串文字可以和應該C. "foo"級聯是一個字符串文字和const char foo[]是一個常數字符串（字符數組）。代碼"foo" "bar"將自動連接，代碼foo bar不會。

如果你願意，你可以隱藏的字符串常量宏背後：

#define foo "foo" 
#define bar "bar" 
char foobar[] = foo bar; // actually works 

So, PLEASE, before you (falsely) claim that C is difficult to use with strings, learn how to use C.

Ç是相當困難與串用，因爲我們可以從這個例子很看。儘管他們的傲慢自信，誰寫的人混在一起的各種概念，仍然必須學習如何使用C.

Answer 2

答案看起來像某人設法將字符串文字（可以用這種方式連接）與const字符串變量進行混合。我的猜測是原來有預處理宏而不是變量。

Answer 3

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

int 
main(int argc, char *argv[]) 
{ 
    char *str1 = "foo"; 
    char *str2 = "bar"; 
    char ccat[strlen(str1)+strlen(str2)+1]; 

    strncpy(&ccat[0], str1, strlen(str1)); 
    strncpy(&ccat[strlen(str1)], str2, strlen(str2)); 
    ccat[strlen(str1)+strlen(str2)+1] = '\0'; 

    puts(str1); 
    puts(str2); 
    puts(ccat); 
} 

此代碼連接str1和str2無需malloc，輸出應該是：

foo 
bar 
foobar