c语言中如何将16进制字符串转化成10进制整数？ 如何用C语言将一个十六进制数字转换为十进制
C\u8bed\u8a00\u7f16\u7a0b\uff08\u5c06\u5341\u516d\u8fdb\u5236\u5b57\u7b26\u4e32\u8f6c\u6362\u4e3a\u5341\u8fdb\u5236\u6574\u6570 \uff09number=0;
k=0;
for(i=0;s[i]!='\0';i++)
{
ch=s[i];
if(ch>='0'&&ch='a'&&ch='A'&&ch<='F')
num[k++]=ch;
}
num[k]='\0'
for(i=0 ;num[i]!='\0',i++)
{
ch=num[i];
if(ch>='0'&&ch<='9')
number=numbe*16+ch-'0';
else if(ch>='a'&&ch<='f')
number=number*16+ch-'a';
else if(ch>='A'&&ch<='F')
number=number*16+ch-'A';
}
\u9700\u8981\u51c6\u5907\u7684\u6750\u6599\u5206\u522b\u6709\uff1a\u7535\u8111\u3001C\u8bed\u8a00\u7f16\u8bd1\u5668\u3002
1\u3001\u9996\u5148\uff0c\u6253\u5f00C\u8bed\u8a00\u7f16\u8bd1\u5668\uff0c\u65b0\u5efa\u4e00\u4e2a\u521d\u59cb.cpp\u6587\u4ef6\uff0c\u4f8b\u5982\uff1atest.cpp\u3002
2\u3001\u5728test.cpp\u6587\u4ef6\u4e2d\uff0c\u8f93\u5165C\u8bed\u8a00\u4ee3\u7801\uff1aint a = 0XFFFF;printf("%d", a);\u3002
3\u3001\u7f16\u8bd1\u5668\u8fd0\u884ctest.cpp\u6587\u4ef6\uff0c\u6b64\u65f6\u6210\u529f\u5c06\u5341\u516d\u8fdb\u5236\u6570\u8f6c\u6362\u4e3a\u5341\u8fdb\u5236\u8fdb\u884c\u4e86\u8f93\u51fa\u3002
3、反转字符串，并赋给字符串数组A
2、循环检测数组A，依次检测1每个字符
5、判断当前检测的字符，如果属于ABCDEF，则依次把ABCDEF 转换为11....16,否则不做任何操作
6、累加，当前循环次数的16次方乘以当前字符的数值大小。
7、输出总累加值就是10进制整数
#include <stdio.h>
#include <stdlib.h>
#include <string.h >
int zh(char s[])
{
int i,m,temp=0,n;
m=strlen(s);//十六进制是按字符串传进来的，所以要获得他的长度
for(i=0;i<m;i++)
{
if(s[i]>='A'&&s[i]<='F')//十六进制还要判断他是不是在A-F或者a-f之间a=10。。
n=s[i]-'A'+10;
else if(s[i]>='a'&&s[i]<='f')
n=s[i]-'a'+10;
else n=s[i]-'0';
temp=temp*16+n;
}
return temp;
}
int main()
{
char s[10];
gets(s);
int n=zh(s);
printf("%d\n",n);
return 0;
}
C语言_012_十六进制转十进制
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