Thread: converting a string to an array

Originally Posted by curlyfries

Sweet, I just got it to work. Look good?

Code:

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

char main(void)
{
char number[5];
printf("Enter a 5 digit number\n");
printf("XXXXX\n");
scanf("%s",number);

int num[5];
num[0]=(number[0]-48);
num[1]=(number[1]-48);
num[2]=(number[2]-48);
num[3]=(number[3]-48);
num[4]=(number[4]-48);

return 0;
}

As for checking to see if the numbers are between 0 and 9, I was thinking of making a simple function that will contain a while loop that checks to see if num[0]-num[4] are between 0 and 9. Will that work?
Dave

Hi Dave,
I see laserlight already gave her critique which is correct as always...

A couple of things...

Move the declaration of num[5] above the first printf. It's more about form than substance but I always like to keep the declarations together and generally I comment what they are for...

Code:

int main(void)
  { char strnum[6];     // number as string from user
     int num[5];        // converted digits
     int x;             // loop counter
   ....

It probably won't make whit of difference for such a small program but when you've written code you have to come back to in 5 years, it makes all the difference in the world. Form good habits early.

Further down you can save yourself some trouble using a loop instead of converting each digit separately...

Code:

for ( x = 0; x < strlen(strnum);x++)
  { num[x] = strnum[x] - '0'; }

Hint: Loops are our friends.

For the range checking...

Code:

for x = 0; x < strlen[number]; x++ )
   { num[x] = number[x] - '0';
       if (num[x] < 0 || num[x] > 9)
         {printf("Digit %d is not acceptable",x);
           num[x] = 0; } }

Ok... you have to know I didn't just hand you the answer...
Find the bugs and let us know how you make out...

Another thing scanf can do is to restrict the input with a regex, so for numeric input restricted to 5 digits this should work.

Code:

scanf("%5[0-9]s", str);

Originally Posted by curlyfries

char main(void) <-- horrible!

Code:

int main()
{
int i,num[5];
printf("Enter a 5 digit number\n");
printf("XXXXX\n");
scanf("%1d%1d%1d%1d%1d",&num[0],&num[1],&num[2],&num[3],&num[4]);
for( i=0;i<5;++i )
  printf("\nnum[%d]=%d %s",i,num[i],num[i]>=0&&num[i]<=9?"OK":"error");
return 0;
}

Originally Posted by BillyTKid

Code:

int main()
{
int i,num[5];
printf("Enter a 5 digit number\n");
printf("XXXXX\n");
scanf("%1d%1d%1d%1d%1d",&num[0],&num[1],&num[2],&num[3],&num[4]);
for( i=0;i<5;++i )
  printf("\nnum[%d]=%d %s",i,num[i],num[i]>=0&&num[i]<=9?"OK":"error");
return 0;
}

I think you missed the point of the exercise...
To convert a numeric string to an array of integers...

I think you missed the point of the exercise...

Originally Posted by BillyTKid

I think you missed the point of the exercise...

this is an example of what I want...

user inputs 14364 as a string (scanf("%s",number))

The number then gets stored in an array like this

int number[5];

number[0]=1
number[1]=4
number[2]=3
number[3]=6
number[4]=4

Nope, don't think so.

Originally Posted by Char*Pntr

Per K & R 2nd ed., "It's bad practice to bury 'magic numbers' in a program; they
convey little information to someone who might have to read the program later, and
they are hard to change in a systematic way."

One suggestion given, in dealing with "magic numbers" is to give them meaningful
names for example, using a #define statement.

I believe the above is the correct concept, not to be taken lightly.

Yep... #define is the way to do things in that case.