怎么样输出二进制不省略前面的零？,C\C++交流,技术交流,鱼C论坛

bz00 发表于 2022-1-21 20:48:05

怎么样输出二进制不省略前面的零？

#include<stdio.h>
#include<stdlib.h>
int main ()
{
int i;
char b;

scanf("%d",&i);
itoa (i,b,2);
printf("%s\n",b);

return 0;
}

比如1的二进制输出结果是1，怎么才能输出00000001这样的形式，谢谢

大马强 发表于 2022-1-21 21:00:02

这样可以吗
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main ()
{
   int i,j,len;
   char b;

   scanf("%d",&i);
   itoa (i,b,2);
   len = strlen(b);
for(j=0;j<(8 -len);j++)
printf("0");
   printf("%s\n",b);


   return 0;
}
3
00000011

--------------------------------
Process exited after 0.5775 seconds with return value 0
请按任意键继续. . .

翼是孤独 发表于 2022-1-21 21:10:06

本帖最后由翼是孤独于 2022-1-22 00:38 编辑

自己写代码补全一下输出字符串就行

第一种方法：
把结果转成字符串，计算一下长度，前面拼接你要的0
比如十进制9，二进制1001，转成字符串4位，你要8位结果，前面就拼接4个0，结果：00001001

第二种方法：
你要8位的结果，原数就加2的8次方即256，转成二进制之后，转字符串，去掉最高位的1
如十进制9，二进制1001，9+256 = 265二进制为 1 0000 1001 ，转成字符串 100001001，去掉第一个1，就是结果

傻眼貓咪 发表于 2022-1-21 22:43:12

#include <stdio.h>

int main()
{
int n, buff = {0};
scanf("%d", &n);
for(int i = 0; n; i++, n /= 2) buff += n%2;
for(int i = 64; i > 0;) printf("%d", buff[--i]);
return 0;
}输入/输出：1234567890
0000000000000000000000000000000001001001100101100000001011010010

bz00 发表于 2022-1-21 23:20:46

大马强发表于 2022-1-21 21:00
这样可以吗

谢谢，看懂了{:5_109:}

bz00 发表于 2022-1-21 23:24:16

翼是孤独发表于 2022-1-21 21:10
自己写代码补全一下输出字符串就行

第一种方法：

第二种方法，能写个代码吗？理论看懂了，刚学习C，还不实现

翼是孤独 发表于 2022-1-22 00:08:16

本帖最后由翼是孤独于 2022-1-22 00:36 编辑

bz00 发表于 2022-1-21 23:24
第二种方法，能写个代码吗？理论看懂了，刚学习C，还不实现

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main ()
{
   int i;
   char b;
   scanf("%d",&i);
   itoa (i+256,b,2);
   strcpy(b,b+1);
   printf("%s\n",b);
   return 0;
}

加一点提示：
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main ()
{
   int i,n;
   char b;
   printf("请输入需要需要转换的数字：");
   scanf("%d",&i);

   printf("请输入最终显示位数：");
   scanf("%d",&n);
   //i + 2^n 转为2进制字符串
   itoa (i+(1<<n),b,2);
   //去掉最高位的1
   strcpy(b,b+1);
   printf("最终结果为：%s\n",b);
   return 0;
}

人造人 发表于 2022-1-22 00:22:57

翼是孤独发表于 2022-1-22 00:08
#include
#include
#include

https://segmentfault.com/q/1010000020075140?utm_source=tag-newest

翼是孤独 发表于 2022-1-22 00:34:32

人造人发表于 2022-1-22 00:22
https://segmentfault.com/q/1010000020075140?utm_source=tag-newest

没问题啊，strycpy(str1,str2) 是把str2 copy到str1
很明显你说的那种溢出错误是因为str2的长度比str1预留的位置大
而这里b+1左移一位后明显比b长度小，所以是不会有问题的

人造人 发表于 2022-1-22 00:38:53

翼是孤独发表于 2022-1-22 00:34
没问题啊，strycpy(str1,str2) 是把str2 copy到str1
很明显你说的那种溢出错误是因为str2的长度比str1预 ...

$ cat main.c
#include <stdio.h>
#include <string.h>

int main(void) {
char str[] = "hello";
strcpy(str, str + 1);
puts(str);
return 0;
}
$ gcc-debug -o main main.c
$ ./main
=================================================================
==606369==ERROR: AddressSanitizer: strcpy-param-overlap: memory ranges [0x7fff4e2911a0,0x7fff4e2911a5) and [0x7fff4e2911a1, 0x7fff4e2911a6) overlap
#0 0x7fc0d796ab37 in __interceptor_strcpy /build/gcc/src/gcc/libsanitizer/asan/asan_interceptors.cpp:438
#1 0x55c7f91f7327 in main /tmp/main.c:6
#2 0x7fc0d6dffb24 in __libc_start_main (/usr/lib/libc.so.6+0x27b24)
#3 0x55c7f91f710d in _start (/tmp/main+0x110d)

Address 0x7fff4e2911a0 is located in stack of thread T0 at offset 32 in frame
#0 0x55c7f91f71e8 in main /tmp/main.c:4

This frame has 1 object(s):
[32, 38) 'str' (line 5) <== Memory access at offset 32 is inside this variable
HINT: this may be a false positive if your program uses some custom stack unwind mechanism, swapcontext or vfork
(longjmp and C++ exceptions *are* supported)
Address 0x7fff4e2911a1 is located in stack of thread T0 at offset 33 in frame
#0 0x55c7f91f71e8 in main /tmp/main.c:4

This frame has 1 object(s):
[32, 38) 'str' (line 5) <== Memory access at offset 33 is inside this variable
HINT: this may be a false positive if your program uses some custom stack unwind mechanism, swapcontext or vfork
(longjmp and C++ exceptions *are* supported)
SUMMARY: AddressSanitizer: strcpy-param-overlap /build/gcc/src/gcc/libsanitizer/asan/asan_interceptors.cpp:438 in __interceptor_strcpy
==606369==ABORTING
$

人造人 发表于 2022-1-22 00:53:18

翼是孤独发表于 2022-1-22 00:34
没问题啊，strycpy(str1,str2) 是把str2 copy到str1
很明显你说的那种溢出错误是因为str2的长度比str1预 ...

$ cat main.c
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

/*
=============
itoa

Convert integer to string

PARAMS:
- value A 64-bit number to convert
- str    Destination buffer; should be 66 characters long for radix2, 24 - radix8, 22 - radix10, 18 - radix16.
- radix Radix must be in range -36 .. 36. Negative values used for signed numbers.
=============
*/

#include <stdbool.h>

char* itoa (unsigned long longvalue,char str[],int radix)
{
char    buf ;
char*    dest = buf + sizeof(buf);
//boolean sign = false;
bool    sign = false;

if (value == 0) {
   memcpy (str, "0", 2);
   return str;
}

if (radix < 0) {
   radix = -radix;
   if ( (long long) value < 0) {
         value = -value;
         sign = true;
   }
}

*--dest = '\0';

switch (radix)
{
case 16:
   while (value) {
         * --dest = '0' + (value & 0xF);
         if (*dest > '9') *dest += 'A' - '9' - 1;
         value >>= 4;
   }
   break;
case 10:
   while (value) {
         *--dest = '0' + (value % 10);
         value /= 10;
   }
   break;

case 8:
   while (value) {
         *--dest = '0' + (value & 7);
         value >>= 3;
   }
   break;

case 2:
   while (value) {
         *--dest = '0' + (value & 1);
         value >>= 1;
   }
   break;

default:          // The slow version, but universal
   while (value) {
         *--dest = '0' + (value % radix);
         if (*dest > '9') *dest += 'A' - '9' - 1;
         value /= radix;
   }
   break;
}

if (sign) *--dest = '-';

memcpy (str, dest, buf +sizeof(buf) - dest);
return str;
}

int main ()
{
   int i,n;
   char b;
   printf("请输入需要需要转换的数字：");
   scanf("%d",&i);

   printf("请输入最终显示位数：");
   scanf("%d",&n);
   //i + 2^n 转为2进制字符串
   itoa (i+(1<<n),b,2);
   //去掉最高位的1
   strcpy(b,b+1);
   printf("最终结果为：%s\n",b);
   return 0;
}
$ gcc-debug -o main main.c
$ ./main
请输入需要需要转换的数字：13
请输入最终显示位数：16
=================================================================
==610718==ERROR: AddressSanitizer: strcpy-param-overlap: memory ranges [0x7ffc5fe90030,0x7ffc5fe90041) and [0x7ffc5fe90031, 0x7ffc5fe90042) overlap
#0 0x7f3976bf9b37 in __interceptor_strcpy /build/gcc/src/gcc/libsanitizer/asan/asan_interceptors.cpp:438
#1 0x5596879c9ea9 in main /tmp/main.c:99
#2 0x7f397608eb24 in __libc_start_main (/usr/lib/libc.so.6+0x27b24)
#3 0x5596879c918d in _start (/tmp/main+0x218d)

Address 0x7ffc5fe90030 is located in stack of thread T0 at offset 80 in frame
#0 0x5596879c9c4d in main /tmp/main.c:88

This frame has 3 object(s):
[48, 52) 'i' (line 89)
[64, 68) 'n' (line 89)
[80, 180) 'b' (line 90) <== Memory access at offset 80 is inside this variable
HINT: this may be a false positive if your program uses some custom stack unwind mechanism, swapcontext or vfork
   (longjmp and C++ exceptions *are* supported)
Address 0x7ffc5fe90031 is located in stack of thread T0 at offset 81 in frame
#0 0x5596879c9c4d in main /tmp/main.c:88

This frame has 3 object(s):
[48, 52) 'i' (line 89)
[64, 68) 'n' (line 89)
[80, 180) 'b' (line 90) <== Memory access at offset 81 is inside this variable
HINT: this may be a false positive if your program uses some custom stack unwind mechanism, swapcontext or vfork
   (longjmp and C++ exceptions *are* supported)
SUMMARY: AddressSanitizer: strcpy-param-overlap /build/gcc/src/gcc/libsanitizer/asan/asan_interceptors.cpp:438 in __interceptor_strcpy
==610718==ABORTING
$

翼是孤独 发表于 2022-1-22 01:12:43

人造人发表于 2022-1-22 00:38

你用的啥编译器啊，刚下了个c编译器试试没报错啊

人造人 发表于 2022-1-22 01:15:11

翼是孤独发表于 2022-1-22 01:12
你用的啥编译器啊，刚下了个c编译器试试没报错啊

带内存检查的gcc
$ alias gcc-debug
alias gcc-debug='gcc -g -Wall -fsanitize=undefined -fsanitize=leak -fsanitize=address -fno-omit-frame-pointer'
$

翼是孤独 发表于 2022-1-22 01:26:21

人造人发表于 2022-1-22 01:15
带内存检查的gcc

老哥不会是竞赛大佬吧，这种内存级别我还是溜了{:10_266:}

人造人 发表于 2022-1-22 01:26:39

翼是孤独发表于 2022-1-22 01:12
你用的啥编译器啊，刚下了个c编译器试试没报错啊

$ cat main.c
#include <stdio.h>
#include <string.h>

int main(void) {
{
   char str[] = "hello";
   strcpy(str, str + 1);
   puts(str);
}
{
   char str[] = "abcde";
   strcpy(str, str + 1);
   puts(str);
}
{
   char str[] = "abcdef";
   strcpy(str, str + 1);
   puts(str);
}
{
   char str[] = "abcd";
   strcpy(str, str + 1);
   puts(str);
}
{
   char str[] = "abc";
   strcpy(str, str + 1);
   puts(str);
}
return 0;
}
$ gcc -g -Wall -o main main.c
$ ./main
eloo
bdee
bceef
bcd
bc
$

人造人 发表于 2022-1-22 01:28:03

翼是孤独发表于 2022-1-22 01:26
老哥不会是竞赛大佬吧，这种内存级别我还是溜了

不是，我自学的
我主修的 C/C++

bz00 发表于 2022-1-22 10:17:59

谢谢各位，都是高手，有些我消化下，有些我还消化不了{:5_109:}

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怎么样输出二进制不省略前面的零？