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1. Write the function int countchtr(char string[ ], int ch);
which returns the number of times the character ch appears
in the string.
Example, the call countchtr(“She lives in NEWYORK”, ‘e’)
would return 3.
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Answer / vignesh1988i
#include<stdio.h>
#include<conio.h>
int countchtr(char [],char);
void main()
{
char a[20],ch;
int c;
printf("enter the string :");
gets(a);
printf("enter the char. to be :");
scanf("%c",&ch);
c=countchtr(a,ch);
printf("%d",c);
getch();
}
int countchtr(char a[],char ch)
{
int count=0;
for(int i=0;a[i]!='\0';i++)
{
if(a[i]==ch)
count++;
}
return(count);
}
thank u
| Is This Answer Correct ? | 4 Yes | 2 No |
Answer / ruchi
#include<stdio.h>
#include<conio.h>
#include<string.h>
int countch(char string[], char );
int main()
{
char str[30],c;
int i=0,s;
printf("\nEnter the string ");
while((str[i++]=getchar())!='\n');
printf("\nEnter the word you want to search ");
scanf("%c",&c);
s = countch(str,c);
if(s !=0)
{
printf("\nTHe total occurence of that word in the string
is %d",s);
}
else
{
printf("\nThe word is not present in the string ");
}
getch();
}
int countch(char str[], char c)
{
int i,sum=0,j,d;
i = strlen(str);
for(j=0;j<i;j++)
{ if(str[j]==c)
{
sum++;
}
}
return (sum);
}
| Is This Answer Correct ? | 2 Yes | 1 No |
#include<stdio.h>
#include<conio.h>
int main()
{
char ptr[100]= "She lives in NEWYORK";
char ch;
printf("ENTER THE CHARACTER:\n");
scanf("%c", &ch);
printf("CHAR %c EXIST %d TIME(S)\n",ch, countchtr(ptr, ch));
getch();
}
int countchtr(char *ptr, char ch)
{
int count = 0;
char ch1;
if(ch >= 97 && ch <= 122)
{
ch1 = ch - 32;
}
else if(ch >= 65 && ch <= 96)
{
ch1 = ch + 32;
}
while(*ptr != '\0')
{ if((*ptr == ch) || (*ptr == ch1))
{
count++;
}
ptr++;
}
return count;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vadivelt
Hi all,
In my post, Answer #3 pls change the statement in if
condition from "ch <= 96" to "ch <= 90"
| Is This Answer Correct ? | 0 Yes | 0 No |
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