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.
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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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