c - 传递 char[] 作为 c 中的引用

Question

我正在尝试将一个 char[] 传递给一个函数并对其进行变异。以下代码针对我的测试用例按预期编译和工作，但我收到了一些编译器警告，我不知道如何解决它们。

代码: ＃包括 #包括

#define MAX_STRING_SIZE 100

void strOp(char **someString) {
  snprintf( someString, sizeof(someString), "HELLO");
}

int main() {
  char myString[MAX_STRING_SIZE];
  strOp(&myString);
  printf("%s\n", myString);
  return 0;
}

警告:

  echoecho.c: In function ‘strOp’:
  echoecho.c:18:3: warning: passing argument 1 of ‘snprintf’ from incompatible pointer type [enabled by default]
     snprintf( someString, sizeof(someString), "HELLO");
     ^
  In file included from echoecho.c:12:0:
  /usr/include/stdio.h:386:12: note: expected ‘char * __restrict__’ but argument is of type ‘char **’
   extern int snprintf (char *__restrict __s, size_t __maxlen,
              ^
  echoecho.c: In function ‘main’:
  echoecho.c:23:3: warning: passing argument 1 of ‘strOp’ from incompatible pointer type [enabled by default]
     strOp(&myString);
     ^
  echoecho.c:17:6: note: expected ‘char **’ but argument is of type ‘char (*)[100]’
   void strOp(char **someString) {

我已经尝试了一些使用 *、*[] 和 ** 的组合，并发现了一些有效但总是有警告。任何想法如何清理它？

Answer 1

将指针传递给char

尝试

#include<cstdlib>
#include<string.h>
#define MAX_STRING_SIZE 100

void strOp(char *someString) {
snprintf( someString,MAX_STRING_SIZE, "HELLO");
}

 int main() {
 char myString[MAX_STRING_SIZE];
 strOp(myString);
 printf("%s\n", myString);
 return 0;  
 }