warning: array ‘str’ assumed to have one element [enabled by default]

[tubby123]

Hi, I get the error "warning: array ‘str’ assumed to have one element [enabled by default]" when I am trying to fill in values inside the string array.

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

char* str[];
int i = 0;

int main(void)
{
 for(i=0;i<5;i++)
 str[i] = (char*)malloc(200 * sizeof(char));
  
 strcpy(str[0],"hello");
 strcpy(str[1],"my"); 
 strcpy(str[2],"name"); 
 strcpy(str[3],"is"); 
 strcpy(str[4],"tubby"); 
 
 for(i=0; i<5; i++)
 printf("[%s]",str[i]);

 return 0;
}

What is the cause for this warning? Yes the output is getting displayed correctly, but why the warning ?
Please note that I have already done a malloc and created a space of 200 characters for each of the strings.

[thines01]

What happens if you give str an initial size of 5?

[tubby123]

What happens if you give str an initial size of 5?

sorry thines01, didn't get you. You mean , what if you give char* str[5] ?

[tubby123]

What happens if you give str an initial size of 5?

@thines01 - you are absolutely right, when I initialize as char* str[5] , it works fine. How is that ? Please explain.

[Ancient Dragon]

char* str[] is the same thing as char** str; , which is a two dimensional array where both dimensions are unspecified and must be allocated at runtime. char* str[5]; is also a two dimensional array, but in this case you have hard-coded one of the dimensions to be of length 5. That means where is room to store 5 strings of unspecified length.

[Adak]

You shouldn't code like that.

Be specific in what you want the compiler to reserve for space.

In fact, if you create an array, and it has almost no space, you can "get away" with overrunning it, by several char's, or pointers to char's, even integers. Usually about 7, but it depends on the compiler, and system.

C does not guarantee that a program with inadequate space will crash - it may run, if you're lucky. Extend this program out to 20 strings, and I'm thinking it's a crasher - but again, I don't code like this, and I've tested it only a few times, with integers.

No, you shouldn't be casting the return from malloc, but most importantly, you should be making your space, adequate and explicit. IMO, this:

char str[n][m] is a true two dimensional array
char *str[n] is an array of pointers to chars, which is very similar to a true 2D array, but not the exact same thing.

Work it clockwise:

str -> [] string is an array
[] -> * of pointers
* - > circles around to the end of the line, then continues on around still going clockwise around str, to find

char

In a spiraling clockwise loop.

[tubby123]

You shouldn't code like that.

Be specific in what you want the compiler to reserve for space.

In fact, if you create an array, and it has almost no space, you can "get away" with overrunning it, by several char's, or pointers to char's, even integers. Usually about 7, but it depends on the compiler, and system.

C does not guarantee that a program with inadequate space will crash - it may run, if you're lucky. Extend this program out to 20 strings, and I'm thinking it's a crasher - but again, I don't code like this, and I've tested it only a few times, with integers.

No, you shouldn't be casting the return from malloc, but most importantly, you should be making your space, adequate and explicit. IMO, this:

char str[n][m] is a true two dimensional array
char *str[n] is an array of pointers to chars, which is very similar to a true 2D array, but not the exact same thing.

Work it clockwise:

str -> [] string is an array
[] -> * of pointers
* - > circles around to the end of the line, then continues on around still going clockwise around str, to find

char

In a spiraling clockwise loop.

@adak - can u maybe rewrite the important parts of the code in the way you would have written. Will help me to understand easier and better.

[Adak]

Hi, I get the error "warning: array ‘str’ assumed to have one element [enabled by default]" when I am trying to fill in values inside the string array.

#include<stdio.h>

int main(void) {
   //unless there's a good reason (like to use global memory), variables are local
   //here only, the compiler will count the number of strings I need, for me
   char *str[]; {"hello","my","name","is","tubby"}; 
   int i = 0;

    for(i=0; i<5; i++)
       printf("%s\n",str[i]);

    return 0;
}

That's how I'd write it.

[rubberman]

Except it should be char *str[] = {"hello","my","name","is","tubby"}; :-)

[Adak]

Yes, only one semi-colon on that line of code! Thanks, Rubberman.