write the function int countchtr(char string[],int
ch);which returns the number of timesthe character ch
appears in the string. for example the call countchtr("she
lives in Newyork",'e') would return 3.
Answers were Sorted based on User's Feedback
Answer / goloap
int countchtr(char *str, char ch)
{
int count=0;
char *itr = str;
while (*itr != '\0')
{
if(*itr == ch)
{
count++
}
itr++;
}
return count;
}
Is This Answer Correct ? | 4 Yes | 0 No |
Answer / vignesh1988i
a small change..............
#include<stdio.h>
#include<conio.h>
int string(char *,char);
void main()
{
char str[100],ch;
int c;
printf("enter the string :");
gets(str);
printf("enter the character to be searched :");
scanf("5c",&ch);
c=string(&str[0],ch);
printf("the character %c occurs for %d times ",ch,c);
getch();
}
int string(char *a,char ch)
{
int count=0;
for(int j=0;*a!='\0';j++)
{
if(*a==ch)
{
count++;
*a++;
}
}
return count;
}
Is This Answer Correct ? | 3 Yes | 0 No |
Answer / vignesh1988i
#include<stdio.h>
#include<conio.h>
int string(char *,char);
void main()
{
char str[100],ch;
int c;
printf("enter the string :");
gets(str);
printf("enter the character to be searched :");
scanf("5c",&ch);
c=string(&str[0],ch);
printf("the character %c occurs for %d times ",ch,c);
getch();
}
int string(char *a,char ch)
{
int count=0;
for(int j=0;*a!='\0';j++)
{
if(*a==ch)
{
count++;
*(a++);
}
}
return count;
}
Is This Answer Correct ? | 4 Yes | 2 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 ? | 1 Yes | 0 No |
Answer / jagjit
#include<stdio.h>
#include<conio.h>
int string(char *,char);
void main()
{
char str[100],ch;
int c;
printf("enter the string :");
gets(str);
printf("enter the character to be searched :");
scanf("5c",&ch);
c=string(&str[0],ch);
printf("the character %c occurs for %d times ",ch,c);
getch();
}
int string(char *a,char ch)
{
int count=0;
for(int j=0;*a!='\0';j++)
{
if(*a==ch)
{
count++;
*(a++);
}
}
return count;
}
Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vadivelt
Hi all,
In my post, Answer #5 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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