Array of pointers and pointer for arrays with argv

Good afternoon. I was wondering why this program with a pointer for arrays with char (*argv)[100] doesn't work at all.

I understand char (*argv)[100] is declaring a pointer for arrays with n lines and a maximum of 13 chars. And char *argv[] is declaring an array for n char pointers (strings). Therefore, I thought the program below would work as expected, printing thames thames thames when I input:

Code:

 
  ./strcpy thames thames thames

My line of thought was: "well, if I input less than 100 characters, everything will work just fine. "

Code:

 
#include <stdio.h> 
#include <stdlib.h> /* for exit and malloc*/
#include <string.h> 

#ifndef MAXPOINTERS
  #define MAXPOINTERS 20
#endif 

#ifndef MAXCHARS
  #define MAXCHARS 100
#endif    

void Strcpy(char*, char*);
char** Malloc2D(int, int);

void Strcpy(char* s, char* t) 
{ 
   while( (*s++ = *t++) )
   ;
}

char** Malloc2D(int nPointers, int nChars)
{
   int i; 
   char** strings;
   
   strings = malloc(nPointers * sizeof(char*));
   
   for(i = 0; i < nPointers; i++) 
   { 
     *(strings + i) = malloc(nChars * sizeof(char));  
   }
   
   return strings;            
}     

int main(int argc, char (*argv)[100]) 
{ 
   int i, j;
   char** s;
   
   
   if(argc < 2) 
   { 
     printf("Please enter [Executable][string]\n");  
     exit(0);
   }
   
   else if(argc >= MAXPOINTERS) 
   { 
     printf("You can write up to %d words", MAXPOINTERS);
     exit(0);   
   }        
   
   s = Malloc2D(MAXPOINTERS, MAXCHARS);
   
   for(j = 1, i = 0; i < argc - 1; i++, j++) 
   { 
      Strcpy(s[i], argv[j]);  
   }        
   
   printf("\nThe strings which were input are:\n\n");
   for(i = 0; i < argc - 1; i++) 
   { 
      printf("%d.\t%s\n", i + 1, s[i]);   
   }         
   
   return 0;
}

When i compile it

Code:

Macintosh-c8bcc88e5669-9:~ usi$ gcc -Wall px.c -o px
px.c:38: warning: second argument of ‘main’ should be ‘char **’

I think here is something you should read.

Originally Posted by std10093

When i compile it

Code:

Macintosh-c8bcc88e5669-9:~ usi$ gcc -Wall px.c -o px
px.c:38: warning: second argument of ‘main’ should be ‘char **’

I think [URL="http://www.cprogramming.com/tutorial/c/lesson14.html"]here/URL] is something you should read.

I knew about this warning, but I'd like to understand why, considering what I said, char (*argv)[50] wouldn't work. will that notation work in another function for what I wanted to print ?

edit:

for example, instead of argv I have char (*str)[50] and I want to print thames

do I have to write char *str[50] to print the output as expected ? why? if I have n letters and I tried to print less than 50 letters?

I think looking up the type would be illuminating.

char (*str)[50]
declare str as pointer to array 50 of char

char *str[50]
declare str as array 50 of pointer to char

So there is a difference, and argv is just a list of pointers at least argc long. And argv[argc] is guaranteed to be NULL.

I think looking up the type would be illuminating

I got the idea.

And argv[argc] is guaranteed to be NULL

this is interesting, I didn't know that. Thanks.