Thread: C program error (need help!)

Thanks!!! Anyway, why don't I need the 'atoi' statement? I have a 'char' and I want to convert it to a 'int'. Could you please explain why I don't need the 'atoi' in this program. Thanks.

I wrote a code to try to understand char/string/int:

#include<stdio.h>

main()
{

/* int c_A,c_a,c_return; */
char c_A, c_a, c_return; /* characters - each 1 byte */

int i;
char c[10]; /* character array - each element has 1 byte*/
/* or can we call it a string? */

c_A='A';
c_a='a';
c_return='\n';

c[0]='0'; c[1]='1'; c[2]='2';
c[3]='3'; c[4]='4'; c[5]='5';
c[6]='6'; c[7]='7'; c[8]='8';
c[9]='9';

for(i=0; i<10; i++) /* first c[i] is promoted to int */
printf("int: %d char: %c \n", c[i], c[i]);

printf("c= %s \n", c); /* also print c as a string */
printf("\n \n");

printf("c_A= %d \n", c_A); /* char promoted to int */
printf("c_a= %d \n", c_a);
printf("c_return= %d \n", c_return);
printf("\n");
printf("c_A repr: %c \n", c_A); /* char as char */
printf("c_a repr: %c \n", c_a);
printf("c_return repr: %c \n", c_return);

return 0;
}

result:

int: 48 char: 0
int: 49 char: 1
int: 50 char: 2
int: 51 char: 3
int: 52 char: 4
int: 53 char: 5
int: 54 char: 6
int: 55 char: 7
int: 56 char: 8
int: 57 char: 9
c= 0123456789


c_A= 65
c_a= 97
c_return= 10

c_A repr: A
c_a repr: a
c_return repr:


Then I use instead:

int i;
char c[]="0123456789";
...

to assign values to c and got the same output. Thus,
is an array of characters such as c[10] defined above also a string? If it's true, then a string is an array holding characters (one char per element), and the value of c[i], i=0,9 can be promoted to be an integer?

Just to test how to use code tags:

Code:

#include<stdio.h>
main()
{
   int i;
   int j;
   i=0; j=1;
         printf("%d %d \n",
                     i,j);
   return 0;
}

How does the "quota tags" work? I saw people using that -- but don't know how it is done.

Code reposted:

Code:

#include<stdio.h>

main()
{ 

/* int c_A,c_a,c_return; */
char c_A, c_a, c_return; /* characters - each 1 byte */

int i;
char c[10]; /* character array - each element has 1 byte*/
/* or can we call it a string? */

c_A='A';
c_a='a';
c_return='\n';

c[0]='0'; c[1]='1'; c[2]='2';
c[3]='3'; c[4]='4'; c[5]='5';
c[6]='6'; c[7]='7'; c[8]='8';
c[9]='9'; 

for(i=0; i<10; i++) /* first c[i] is promoted to int */
printf("int: %d char: %c \n", c[i], c[i]);

printf("c= %s \n", c); /* also print c as a string */
printf("\n \n");

printf("c_A= %d \n", c_A); /* char promoted to int */
printf("c_a= %d \n", c_a);
printf("c_return= %d \n", c_return); 
printf("\n");
printf("c_A repr: %c \n", c_A); /* char as char */
printf("c_a repr: %c \n", c_a);
printf("c_return repr: %c \n", c_return);

return 0;
}

result:

int: 48 char: 0
int: 49 char: 1
int: 50 char: 2
int: 51 char: 3
int: 52 char: 4
int: 53 char: 5
int: 54 char: 6
int: 55 char: 7
int: 56 char: 8
int: 57 char: 9
c= 0123456789


c_A= 65
c_a= 97
c_return= 10

c_A repr: A
c_a repr: a
c_return repr:


Then I use instead:

int i;
char c[]="0123456789";
...

to assign values to c and got the same output. Thus,
is an array of characters such as c[10] defined above also a string? If it's true, then a string is an array holding characters (one char per element), and the value of c[i], i=0,9 can be promoted to be an integer?

Sorry... the tab key accidetally submits the last post....

Code:

#include<stdio.h>

main()
{ 

       /* int c_A,c_a,c_return; */
        char c_A, c_a, c_return;        /* characters - each 1 byte */

        int i;
        char c[10];     /* character array - each element has 1 byte*/
                                  /* or can we call it a string? */

        c_A='A';
        c_a='a';
        c_return='\n';

         c[0]='0'; c[1]='1'; c[2]='2';
         c[3]='3'; c[4]='4'; c[5]='5';
         c[6]='6'; c[7]='7'; c[8]='8';
         c[9]='9'; 

               for(i=0; i<10; i++)        /* first c[i] is promoted to int */
                      printf("int: %d char: %c \n", c[i], c[i]);

          printf("c= %s \n", c);           /* also print c as a string */
          printf("\n \n");

         printf("c_A= %d \n", c_A);      /* char promoted to int */
         printf("c_a= %d \n", c_a);
         printf("c_return= %d \n", c_return); 

         printf("\n");
         printf("c_A repr: %c \n", c_A); /* char as char */
         printf("c_a repr: %c \n", c_a);
         printf("c_return repr: %c \n", c_return);

          return 0;
}

result:

int: 48 char: 0
int: 49 char: 1
int: 50 char: 2
int: 51 char: 3
int: 52 char: 4
int: 53 char: 5
int: 54 char: 6
int: 55 char: 7
int: 56 char: 8
int: 57 char: 9
c= 0123456789


c_A= 65
c_a= 97
c_return= 10

c_A repr: A
c_a repr: a
c_return repr:


Then I use instead:

int i;
char c[]="0123456789";
...

[quota]
to assign values to c and got the same output. Thus,
is an array of characters such as c[10] defined above also a string? If it's true, then a string is an array holding characters (one char per element), and the value of c[i], i=0,9 can be promoted to be an integer?
[/quota]