我正在尝试计算 C 中字符串中的数字

Question

What's wrong with my code? I'm trying to count the numbers in a string after skipping all alphabets.

How to skip the alphabets and only count the numbers?

#include <stdio.h>
#include <stdlib.h>

void function(char a[])
{
    int i=0,count=0;
    while(a[i]!='\0')
    {
        if(a[i]>='a'&&a[i]<='z')
        {
            continue;
        }
        else
        {
            count++;
        }
        i++;
    }
    printf("%d",count);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char a[100001];
        scanf("%s",a);
        function(a);
        printf("\n");
    }
}

Answer 1

I switched your usage of scanf to use fgets . This works:

#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>

void function(char a[])
{
    int i=0,count=0;
    while(a[i]!='\0')
    {
        if (isdigit(a[i]))
        {
            count++;
        }
        i++;
    }
    printf("%d",count);
}

int main()
{
    int t = 0;

    scanf("%d", &t);

    while(t--)
    {
        char a[100001];
        fgets(a, sizeof(a), stdin);
        function(a);
        puts("\n");
    }
}

Answer 2

If the OP's intent was in fact to count up the groups of digits (in which the definition of 'number' is each contiguous run of digits), this will accomplish the goal:

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

int function(char a[])
{
    size_t i = 0;
    int cnt = 0;
    //  convert all non-digits to whitespace
    printf("Converting\n");
    for (char *p = a; *p != '\0'; ++p)
    {
        *p = isdigit(*p) ? *p : ' ';
    }

    printf("Found Groups: %s\n",a);
    // note: strtok mutates the string it is given, it will not be usable after this:
    for (char *gr = strtok(a," "); gr != NULL; gr = strtok(NULL, " "))
    {
        cnt++;
    }
    printf("Number of Groups: %d\n", cnt);

}
int main()
{
    printf("Enter strings (Ctrl-C to end)\n");
    while(1)
    {
        int result = 0;
        char a[100001];
        scanf("%s",a);
        result = function(a);
        printf("\n");
    }
}

(I was not able to get the version using fgets to work, but how the string is gathered is not really in the scope of the question, the OP's original code in main is functional, I just made it an open-ended loop for my tests)

Answer 3

In your code continue statement is responsible for wrong answer. Because of it your code is not encountering the statement a[i].='\0'. so suggest you to just avoid it and close your if statement without any code as like me. One more suggestion use also uppercase letters as user can also enter uppercase letters and using fgets() function you can also read white spaces. Try out this

#include <stdio.h>
#include <stdlib.h>

void function(char a[])
{
   int i=0,count=0;
  while(a[i]!='\0')
 {
        if((a[i]>='a'&&a[i]<='z') || (a[i]>='A'&&a[i]<='Z'));
       else
      {
         count++;
        }
        i++;
    }
    printf("%d",count);
}

int main()
{
   int t;
  scanf("%d",&t);
 while(t--)
    {
       char a[100001];
      scanf("%s",a);
     function(a);
        printf("\n");
    }
}