#include "Stdio.h"
#include "Conio.h"
#include "String.h"
#include "math.h"
int main(void)
{
int transform(char str[100],int length);
int length,i;
char str[100];
long number;
gets(str);
length=strlen(str);
for(i=0;i<=length;i++)
{
if(str[i]>=97&&str[i]<=102)str[i]=str[i]-39;
}
number=transform(str,length);
printf("number=%d",number);
getch();
return 0;
}
int transform(char str[100],int length)
{
int i,k=0;
long number=0;
for(i=length;i>=0;i--)
{
number=number+pow(16,k)*(str[i]);
k++;
}
return number;
}
主要问题就在于16进制中ABCDE中应该怎么转化，另外字符数组中用于计算如 number=number+pow(16,k)*(str[i]);中str[i]是指其所指向的值还是ASCII值……问题比较简单但是这个程序编了一个小时还是不明白哪里错了……
#include<stdio.h>
int main()
{
int n;
char *str = "ABCDE";
sscanf(str, "%X", &n);
printf("%d\n", n);
return 0;
}
先有个判断
如果是ABCDEF这六个可以str[i] - ‘A' + 10
如果是0123456789这十个可以str[i] - ’0‘追问
恩……能不能不用sscanf函数……另外我ABCDEF改了也不行，还有我们要求是小写字母
追答小写字母就是str[i] - ‘a' + 10
追问是这么 改的，但是VS2010运行各种不对……
追答#include "Stdio.h"
#include "Conio.h"
#include "String.h"
#include "math.h"
int main(void)
{
int transform(char str[100],int length);
int length,i;
char str[100];
long number;
gets(str);
length=strlen(str);
number=transform(str,length);
printf("number=%d",number);
getch();
return 0;
}
int transform(char str[100],int length)
{
int i,k=0;
long number=0;
for(i=length-1;i>=0;i--)
{
int tmp = 0;
if('0'<= str[i] && str[i] <= '9') tmp = str[i] - '0';
if('A'<= str[i] && str[i] <= 'F') tmp = str[i] - 'A' + 10;
if('a'<= str[i] && str[i] <= 'f') tmp = str[i] - 'a' + 10;
number += pow(16, k) * tmp;
k++;
}
return number;
}
#include "Conio.h"
#include "String.h"
#include "math.h"
int main(void)
{
int transform(char str[100],int length);
int length,i;
char str[100];
long number;
gets(str);
length=strlen(str);
number=transform(str,length);
printf("number=%d",number);
getch();
return 0;
}
int transform(char str[100],int length)
{
int i,k=0;
long number=0;
for(i=length-1;i>=0;i--)
{
int tmp = 0;
if('0'<= str[i] && str[i] <= '9') tmp = str[i] - '0';
if('A'<= str[i] && str[i] <= 'F') tmp = str[i] - 'A' + 10;
if('a'<= str[i] && str[i] <= 'f') tmp = str[i] - 'a' + 10;
number += pow(16, k) * tmp;
k++;
}
return number;
}
