Strings - need digits

This is a discussion on Strings - need digits within the C Programming forums, part of the General Programming Boards category; Hi I am trying to write a code that accpets a string of characters. I need to count the number ...

  1. #1
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    Question Strings - need digits

    Hi

    I am trying to write a code that accpets a string of characters. I need to count the number of digits in that string and return the counted value.

    This is my code
    include <stdio.h>
    #include <ctype.h>

    #define MAX 15

    int digcnt(int c);

    int main( int argc, char *argv[]) {
    char x[MAX];
    int c;

    printf("Enter a string of chars and digits:\n");

    // fgets(x, MAX, stdin);
    while(( c= getchar()) !='\n') {
    c =x[MAX];

    }

    digcnt(c);
    return 0;
    }

    int digcnt(int c) {

    int b;

    for(b=0; b<MAX; b++) {
    if(isdigit(c)){
    b++;
    }
    }
    printf("There are %d digits in the string.\n", b);
    return b;
    }

    Also how do i check if the string (in my case, x[MAX]) contains digits?
    My output is that it gives me the number of MAX as the number of digits.

    All help is appreciated.
    Last edited by s.sidak; 10-26-2010 at 11:34 PM.

  2. #2
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    i think u should use sscanf(x, "%d", &c);
    inside if to check if the c is a digit
    like if ( x[i] >= '0' && x[i] <= '9')
    i'm not sure

  3. #3
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    Quote Originally Posted by s.sidak View Post
    Hi

    I am trying to write a code that accpets a string of characters. I need to count the number of digits in that string and return the counted value.
    All help is appreciated.
    Ok, there are a number of problems in the sample provided but most seem to follow the error shown here...
    Code:
    // fgets(x, MAX, stdin);
    while(( c= getchar()) !='\n') {
    c =x[MAX];
    Think about what you are assigning to what... The goal is to build a string with the typed characters in it... is it not? Which is your string and which is your input variable?
    How do you add new characters to the string?

    In C results are always on the left. It's not 6 + 4 = 10 It's 10 = 6 + 4

  4. #4
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    Hi,
    a small program to do what you need :

    Code:
    #include <stdio.h>
    #include <ctype.h>
    
    
    #define		BUFFER	16
    
    int
    main(int argc, char **argv)
    {
    	int		count = 0;
    	char	my_string[BUFFER];
    	int		i;
    	
    	while (fgets(my_string, BUFFER, stdin)) {
    		for(i = 0; i < BUFFER; i++) {
    			if (isdigit(my_string[i]))
    				count++;
    			if (my_string[i] == '\n')
    				goto end;
    		}
    	}
    end:
    	printf("Digits : %d\n", count);
    	
    	return 0;
    }
    Regards.

  5. #5
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    Quote Originally Posted by evariste View Post
    Hi,
    a small program to do what you need :
    And how much do you think our friend learned from that?

    I don't mean to be rude but people don't learn when you simply hand them the answers.

  6. #6
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    fgets is better than getchar/scanf("%c") stuff.
    You need another variable for count the digits, you use one variable for iteration and count.
    Code:
    #include <stdio.h>
    #include <ctype.h>
    
    #define MAX 15
    
    int digcnt(char *x);
    
    int main( int argc, char *argv[]) {
    char x[MAX];
    
    puts("Enter a string of chars and digits:");
    
    fgets(x, MAX, stdin);
    
    digcnt(x);
    return 0;
    }
    
    int digcnt(char *x) {
    
    int b,count=0;
    
    for(b=0; x[b]!='\0'; b++) {
    if(isdigit(x[b])){
    count++;
    }
    }
    printf("There are %d digits in the string.\n", count);
    return count;
    }

  7. #7
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    Quote Originally Posted by CommonTater View Post
    And how much do you think our friend learned from that?

    I don't mean to be rude but people don't learn when you simply hand them the answers.
    That's true , i'm sorry.

    Regards.

  8. #8
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    Hi all

    Thank You very much for ur replies. Everything is working now, all thanks to your help and by showing me the examples.

    I was able to understand the two codes given.

    Thanks for your help.

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