一个c语言的小疑点,
一个c语言的小问题,请教高手。。。
给定一个指向B个字节的指针p,如何才能将这B个字节所包含的01串,转化为一个无符号整数呢?
------解决方案--------------------
仅供参考
------解决方案--------------------
我想这样应该可以,在x86上不考虑bus error。
------解决方案--------------------
可以的。
或者
给定一个指向B个字节的指针p,如何才能将这B个字节所包含的01串,转化为一个无符号整数呢?
------解决方案--------------------
仅供参考
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <locale.h>
int main() {
int i,v;
char bs[33];
char b[33];
char hs[9];
char h[9];
char s[4];
char *e;
// 十进制整数转二进制串;
i=1024;
ltoa(i,b,2);
sprintf(bs,"%032s",b);
printf("i=%d,bs=%s\n",i,bs);
// 十进制整数转十六进制串;
i=1024;
ltoa(i,h,16);
sprintf(hs,"%08s",h);
printf("i=%d,hs=%s\n",i,hs);
// 十六进制字符串转成十进制数
strcpy(hs,"00000400");
sscanf(hs,"%x",&i);
printf("hs=%s,i=%d\n",hs,i);
// 二进制字符串转化为十六进制字符串;
strcpy(bs,"00000000000000000000010000000000");
i=strtol(bs,&e,2);
ltoa(i,h,16);
sprintf(hs,"%08s",h);
printf("bs=%s,hs=%s\n",bs,hs);
// 二进制字符串转化为十进制数;
strcpy(bs,"00000000000000000000010000000000");
i=strtol(bs,&e,2);
printf("bs=%s,i=%d\n",bs,i);
// 十六进制字符串转成二进制串
strcpy(hs,"00000400");
sscanf(hs,"%x",&i);
ltoa(i,b,2);
sprintf(bs,"%032s",b);
printf("hs=%s,bs=%s\n",hs,bs);
// ASC\GBK字符串转十六进制串
strcpy(s,"a汉");
i=0;
while (1) {
if (0==s[i]) break;
sprintf(hs+i*2,"%02X",(unsigned char)s[i]);
i++;
}
setlocale(LC_ALL,"chs");
printf("s=%s,hs=%s\n",s,hs);
// 十六进制字符串转成汉字(GBK)及字符(ASC)
strcpy(hs,"61BABA");
i=0;
while (1) {
if (1!=sscanf(hs+i*2,"%2x",&v)) break;
s[i]=(char)v;
i++;
}
s[i]=0;
printf("hs=%s,s=%s\n",hs,s);
return 0;
}
//i=1024,bs=00000000000000000000010000000000
//i=1024,hs=00000400
//hs=00000400,i=1024
//bs=00000000000000000000010000000000,hs=00000400
//bs=00000000000000000000010000000000,i=1024
//hs=00000400,bs=00000000000000000000010000000000
//s=a汉,hs=61BABA
//hs=61BABA,s=a汉
------解决方案--------------------
我想这样应该可以,在x86上不考虑bus error。
unsigned int *q = p; //(p为char *)
cnt = B / sizeof(unsigned int);
int i;
for(i = 0;i < cnt;i++)
{
printf("%u\n",*q);
q++;
}
p = q;
int rest = B - cnt*szieof(unsigned int);
switch(rest)
{
case 1:
printf("%u\n",(unsigned int)*p);break;
case 2:
printf("%u\n",(unsigned int)(*p
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(*(p+1) << 8));break;//小端模式下
case 3:
printf("%u\n",(unsigned int)(*p
------解决方案--------------------
(*(p+1) << 8)
------解决方案--------------------
(*(p+2) << 16));break;
default:
break;
}
------解决方案--------------------
可以的。
unsigned char* p = ...;
unsigned long long v = 0;
for(size_t i = 0; i < B; ++i){
v <<= 8;
v += *p;
}
或者