Write code for atoi(x) where x is hexadecimal string.
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Answer / john huang
int n=strlen(x) // where x is pointer to hex string
int sum=0;
int leftshift=0;
while(n>0)
{
if((x[n-1]>='0') && (x[n-1]<='9'))
sum+=(x[n-1]-'0')<<leftshift;
if((x[n-1]>='A') && (x[n-1]<='F'))
sum+=(x[n-1]-'A'+10)<<leftshift;
if((x[n-1]>='a') && (x[n-1]<='f'))
sum+=(x[n-1]-'a'+10)<<leftshift;
n--;
leftshift+=4;
}
| Is This Answer Correct ? | 7 Yes | 3 No |
Answer / mohammed sardar
int n=strlen(x) // where x is pointer to hex string
int sum=0;
int leftshift=0;
while(n>0)
{
if((x[n-1]>='0') && (x[n-1]<='9'))
sum+=(x[n-1]-'0')<<leftshift;
if((x[n-1]>='A') && (x[n-1]<='F'))
sum+=(x[n-1]-'A'+10)<<leftshift;
if((x[n-1]>='f') && (x[n-1]<='f'))
sum+=(x[n-1]-'a'+10)<<leftshift;
n--;
leftshift+=4;
}
| Is This Answer Correct ? | 10 Yes | 7 No |
Hi,
Refer below link to know how atoi() lib fuction works.
http://www.cppreference.com/wiki/c/string/atoi
And find the equalent code which i have written here.
#include<stdio.h>
#include<conio.h>
int MyAtoi(char *cptr);
main()
{
/*Give different inputs like "12.3432", "a4523"," 123"
"abcd", "1234f" and find the qualent output*/
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;
}
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / audrius
long l = strtol(string, NULL, 16);
//or, if you need unsigned (which you usually do with hexes)
unsigned long ul = strtoul(string, NULL, 16);
| Is This Answer Correct ? | 1 Yes | 0 No |
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