Question

这里我创建了一个字符串，我在字符串中存储了数字的二进制值..我想将变量num的值存储到字符串中。

我包含给定十进制数的二进制数的长度。假设给定的数字是A = 6，我包含3，我需要一个字符串＆＃39;结果＆＃39;有＆＃39; 110＆＃39;这是二进制值6.

char* result = (char *)malloc((i)* sizeof(char));
i--;
while(A>=1)
    {
        num=A%2; 
        result[i]=num;  // here I need to store the value of num in the string
        A=A/2;
        i--;
     }

Answer 1

It appears from the code you've posted is that what you are trying to do is to print a number in binary in a fixed precision. Assuming that's what you want to do, something like

unsigned int mask = 1 << (i - 1);     
unsigned int pos = 0;
while (mask != 0) {
    result[pos] = (A & mask) == 0 ? '0' : '1';
    ++pos;
    mask >>= 1;
}
result[pos] = 0; //If you need a null terminated string

edge cases left as an exercise for the reader.

Answer 2

I'm not sure specifically what you are asking for. Do you mean the binary representation (i.e. 00001000) of a number written into a string or converting the variable to a string (i.e. 8)? I'll assume you mean the first.

The easiest way to do this is to repeatedly test the least significant bit and shift the value to the right (>>). We can do this in for loop. However you will need to know how many bits you need to read. We can do this with sizeof.

int i = 15;
for (int b = 0; b < sizeof(i); ++b) {
    uint8_t bit_value = (i & 0x1);
    i >>= 1;
}

So how do we turn this iteration into a string? We need to construct the string in reverse. We know how many bits are needed, so we can create a string buffer accordingly with an extra byte for NULL termination.

char *buffer = calloc(sizeof(i) + 1, sizeof(char));

What this does is allocates memory that is sizeof(i) + 1 elements long where each element is sizeof(char), and then zero's each element. Now lets put the bits into the string.

for (int b = 0; b < sizeof(i); ++b) {
    uint8_t bit_value = (i & 0x1);

    size_t offset = sizeof(i) - 1 - b;
    buffer[offset] = '0' + bit_value;

    i >>= 1;
}

So what's happening here? In each pass we're calculating the offset in the buffer that we should be writing a value to, and then we're adding the ASCII value of 0 to bit_value as we write it into the buffer.

This code is untested and may have some issues, but that is left as an exercise to the reader. If you have any questions, let me know!

Answer 3

这是整个代码。它应该工作正常。

int i=0;
int A;//supposed entered by user

//calculating the value of i
while(A!=0)
{
    A=A/2;
    i++;
}
char* result=(char *)malloc(sizeof(char)*i);
i--;
while(A!=0)
{
    result[i]='0'+(A%2);
    A=A/2;
    i--;
}

Answer 4

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

char *numToBinStr(int num){
    static char bin[sizeof(int) * CHAR_BIT + 1];
    char *p = &bin[sizeof(int) * CHAR_BIT];//p point to end
    unsigned A = (unsigned)num;

    do {
        *--p = '0' + (A & 1);
        A >>= 1;
    }while(A > 0);//do-while for case value of A is 0 
    return p;
}

int main(void){
    printf("%s\n", numToBinStr(6));
    //To duplicate, if necessary
    //char *bin = strdup(numToBinStr(6));
    char *result = numToBinStr(6);
    char *bin = malloc(strlen(result) + 1);
    strcpy(bin, result);
    printf("%s\n", bin);
    free(bin);
    return 0;
}

Answer 5

您可以在<stdlib.h>：

中使用这些功能

itoa();或sprintf()

第二个链接也有一些例子。

如何将变量的值存储到字符串数组？

5 个答案: