如何修复函数及其声明的“冲突类型”？

Question

I want to pass a two-dimensional char array to a function but don't know how to declare the function before the main(). The function compiles and works well before I declare it. But after I declare it, I encounter compiling issues.

I'm using EMACS on MacBook pro. The compiler is gcc.I tried to declare my function print string various ways including

void printstring(int, int,char **);

or

void printstring(int, int,char *); 

But none of them work. My Full codes are:

#include<stdio.h>

#include<stdlib.h>

void printstring(int, int,char **);

int main(){
  char word[3][6]= {"hello","world","I"};
  printstring(3,6,word);
  return 0;
}

void printstring(int n, int m, char (*w)[m]){
  for (int i = 0; i < n; i++){
    printf("%s\n",w[i]);
  }
  return;
}

I expected that there is no compiling error but I got one error and one warning. Details can be found below:

test.c: In function 'main':
test.c:9:19: warning: passing argument 3 of 'printstring' from incompatible pointer type [-Wincompatible-pointer-types]
   printstring(3,6,word);
                   ^~~~
test.c:5:6: note: expected 'char **' but argument is of type 'char (*)[6]'
 void printstring(int, int,char **);
      ^~~~~~~~~~~
test.c: At top level:
test.c:13:6: error: conflicting types for 'printstring'
 void printstring(int n, int m, char (*w)[m]){
      ^~~~~~~~~~~
test.c:5:6: note: previous declaration of 'printstring' was here
 void printstring(int, int,char **);
      ^~~~~~~~~~~

Answer 1

the problem is that you're using a variable length array. The last argument (the list of strings) depends on the second argument ( m ). And char ** is not suitable, as it's just a pointer on pointers. So the max dimension of the strings would be lost when iterating on the 2D array.

Use a standard forward declaration, copying exactly the real declaration if you don't want to put the function before the main one.

void printstring(int n, int m, char (*w)[m]);

int main(){
  char word[3][6]= {"hello","world","I"};
  printstring(3,6,word);
  return 0;
}
void printstring(int n, int m, char (*w)[m]){
  for (int i = 0; i < n; i++){
    printf("%s\n",w[i]);
  }
  return;
}

If you have read-only strings, I suggest that you use a standard array of constant pointers instead:

void printstring(int n, const char *w[]);

int main(){
  const char *word[] = {"hello","world","I"};
  printstring(3,word);
  return 0;
}
void printstring(int n, const char *w[])
{
  for (int i = 0; i < n; i++){
    printf("%s\n",w[i]);
  }
  return;
}

note that

printstring(3,word);

can be replaced by

printstring(sizeof(word)/sizeof(word[0]),word);

before array decays to pointer (that autocomputes the number of strings)

Answer 2

The following should just work:

void printstring(int n, int m, char (*w)[m]);

The function prototype and definition should be kept identical, except maybe for certain qualifiers such as const and default arguments in C++.

Answer 3

If you want to maintain a name free declaration for whatever reason, you can use the * notation (reserved to function prototype scope) for the variably modified type

void printstring(int, int,char (*)[*]);

Still a VLA, and in fact, exactly equivalent to the notation that uses m . Though, ostensibly, it may convey intent not as clearly as using m in the forward declaration.

Answer 4

char** cannot be used to point at 2D arrays, it can only be used to point at the first element of a 1D array of char* , which is something else.

Your compiler error is from having non-matching declaration and definition. Correct code:

void printstring(int n, int m, char w[n][m]);

...

void printstring(int n, int m, char w[n][m]){
  ...
}

Alternatively, you can write void printstring(int n, int m, char (*w)[m]) and it is completely equivalent. But that is just harder to read, so why would you?