Thread: return string should work, but ..

Hello,

i am new to C, but know some basic and z80source
i try to make a decimal-string from any 8bit value, so string from 3 chars + \0
it all works, but i always have the wrong type of 'return' value
if it realy works, i remove all those printstatements.
version 038 by now:

Code:

#include <stdio.h>

char * strint(int num ) {
/* avoiding other libs , keeping it simple */

if (num >255) return ;  /* set 8bit maximum */

char str[3] ;  /* '255'  needs 3 char spaces */

int s ;

for (s=0 ; s<3 ; s++) str[s]=' ' ;
str[3]='\0';

printf("1-%d..%s..\n",num, str);

str[0] = num /100 ;
printf("2-%c..\n",str[0]+48);

num = num -100* str[0];
str[0]=str[0]+48  ; /*ascii*/
printf("3-%d..%s..\n",num, str);

str[1] = num/10 ;
printf("4-%c..\n",str[1]+48);

num = num -10* str[1];
str[1]=str[1]+48  ; /*ascii*/
printf("5-%d..%s..\n",num, str);

str[2]=num;
printf("6-%c..\n",str[2]+48);
str[2]=str[2]+48  ; /*ascii*/
printf("7-%d..%s..\n",num, str);

 return  *str ;
  
}

void main() {
  printf( "the number is: %c \n" , strint(213) ) ;
}

result:
ch@linux-4xd4:~/Downloads/fuse-1.3.3> ./strint_038
1-213..   ..
2-2..
3-13..2  ..
4-1..
5-3..21 ..
6-3..
7-3..213..
the number is: 2

printf( "the number is: %s \n" , strint(213) ) ;

gives
1-213.. ..
2-2..
3-13..2 ..
4-1..
5-3..21 ..
6-3..
7-3..213..
Segmentation fault

but it is a string, it should be,
what do i mis here?
i guess its simple, hope so anyway

Chris

Originally Posted by algorism

Can anyone actually read this post?
I can't see it at all.
This has happened before.
What's going on?

I can read this post; but, it is the only one in the thread.

Tim S.

> with out this printf-statement there is only garbage
Because
> warning: function returns address of local variable

The compiler isn't lying.
If you return a pointer to a local variable, the variable goes out of scope with the return.
By going out of scope, the associated memory is free to be used by something else instead.


A crude way to fix the problem is to make the memory static, like so

Code:

char * strint(int num ) {
/* avoiding other libs , keeping it simple */
if (num >255) return NULL ;  /* set 8bit maximum */

static char str[4] ;  /* '255\0'  needs 4 char spaces */
...

But this has the problem that you can only call the function serially.
This would work.
printf( "the number is: %s \n" , strint(213) ) ;
This wouldn't.
printf( "the numbers are: %s %s \n" , strint(213) , strint(42) ) ;


A better way (say like sprintf) is to pass the buffer you want to store the string in.

Code:

char * strint(int num, char *str) {
...
    return str;
}

int main ( ) {
  char buff[4];
  printf( "the number is: %s \n" , strint(213,buff) ) ;
}

> char str[3] ; /* '255' needs 3 char spaces */
Except you always put a \0 on the end, so you really need 4 characters.



Also, enable as many warnings as possible.

Code:

$ gcc -Wall -Wextra foo.c
foo.c: In function ‘strint’:
foo.c:21:8: warning: function returns address of local variable [-Wreturn-local-addr]
 return str ;
        ^
foo.c: At top level:
foo.c:24:6: warning: return type of ‘main’ is not ‘int’ [-Wmain]
 void main() {
      ^

the last and working version

Code:

#include <stdio.h>

char * strint(int num ) {
/* avoiding other libs , keeping it simple */

static char str[3] ;  /* '255'  needs 3 char spaces */
str[3]='\0';  

if (num <0 || num >255) { str[0]='n'; str[1]='o'; str[2]='p'; goto back ; } /* set 8bit maximum */

int s, dv ;

dv=100;
for (s=0 ; s<3 ; s++) {
str[s] = num /dv ;
num -= dv * str[s];
str[s]+= 48  ; /*ascii*/
dv= dv/10 ;
}

back:
return str ;
}

int main() {
  int temp ;
  printf("give any number from 0 to 255\n");
  scanf("%d", &temp);
  printf( "the number is: %s \n" , strint(temp) ) ;
}

> the last and working version
Nope - you still have an array overrun.

You should fix your 'goto' with an if / else construct.

If you wrote
str[s]+= '0';
you wouldn't need the comment, and you wouldn't be assuming that the world uses ascii.

Originally Posted by Satya

In c array indexing starts from 0. Hence if it is a[3]. You can have a[0],a[1],a[2] only and not a[3].

I counted something dubbel.
quote"char string[50];This would declare a string with a length of 50 characters." declaring is the AMOUNT and the first is 0.
In zx basic there is no 0-index in arrays.
So after the remark to include 0-index, my mind made it as if [3] is the place AND amount with that automatic '\0' somewhere
but the declaration is the amount and there is the "default 0+1 error" very much available, so to say.
if i need 4 char, then i declare 4 char.
and apperently the '\0' marker is not needed, since its not used, i think,. i guess,...
but if i use a array part that does NOT exsist, then Why is it accepted?
is it created after all ???

In order to store a string properly you must count all of the characters you will use: you can see that "255" is 3 characters but you should always add 1 for the terminating zero ('\0'). If you fail to do that or make some other mistake with strings, the extra space required is not created for you. C is not like newer languages where arrays are checked and errors always occur.

I will show you how an array overrun does not necessarily lead to an obvious error, since I think that is the best way to explain.

Code:

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

int main(void)
{
   struct strings_type 
   {
      char foo[5];
      char bar[6];
   } example;

   strcpy(example.foo, "0123456789"); /* string is deliberately too long - needs 11 */
   puts(example.foo);
   return 0;
}
 
0123456789

The strcpy() function wrote into other nearby memory, bar, which is clearly wrong, but it executes just fine.

Now this code is easy to fix by using strncpy() because that function will only copy a set number of characters into the array, but you will have to be similarly careful with how you manipulate strings.