Answer / ramu

int f(char s[])
{
int num=0,i;
for(int i=0;s[i]>='0' && s[i]<='9';i++)
n=10*n+(s[i]-'0');
if(i<sizeof(s))
{
printf("Error String");
return 0;
}
return n;
{

Answer / swapnil chhajer

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

int main()
{
char str[10];
printf("Enter the string : ");
gets(str);
printf("Converted integer : %d",atoi(str));
getchar();
}

Answer / vadivel t

The equalent code to atoi() library fuction which i hav
written, below.


#include<stdio.h>
#include<conio.h>

int MyAtoi(char *cptr);

main()
{
char *cptr = "123445";

printf("INTEGER EQU IS: %d\n", MyAtoi(cptr));
getch();
}
int MyAtoi(char *cptr)
{
int iptr = 0;
while((*cptr != '\0') && ((*cptr >= 48 && *cptr <= 57) ||
(*cptr == 32)))
{
if(*cptr != ' ')
iptr = (iptr * 10) + (*cptr - 48);
cptr++;
}
return iptr;
}

Answer / vignesh1988i

#include<stdio.h>
#include<conio.h>
#include<string.h>
int fun(char *,int);
void main()
{
long int c,l;
char a1[20];
printf("enter the numerical string :");
scanf("%s",&a1);
l=strlen(a1);
c=fun(a,l);
printf("\n%ld",l);
getch();
}
int fun(char *a,long int l1)
{
long int a1[]=
{1,10,100,1000,10000,100000,1000000,10000000,100000000},c;
int p=a1[l1-1],c=0;
for(int i=0;a[i]!='\0';i++)
{
y=((int)a[i])*p; /*TYPE CASTING*/
c=c+y;
p=p/10;
}
return c;
}

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