将十六进制转换为二进制

Question

我试图弄清楚为什么这个代码只用2位数的十六进制数字。例如，如果输入“11”，它将输出“00010001”，但如果输入“111”，则会给我一些随机数。我想尽量让它接受用户想要的数字。

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

void binary_hex(int n, char hex[]); 
int hex_binary(char hex[]); 

int main() 
{ 
    char hex[20],c; 
    int n; 

    printf("Enter hexadecimal number: "); 
    scanf("%s",hex); 
    printf("Binary number: %d",hex_binary(hex)); 

    system("pause"); 
    return 0;} 

//Function to convert hexadecimal to binary. 

int hex_binary(char hex[]) { 

int i, length, decimal=0, binary=0; 
for(length=0; hex[length]!='\0'; ++length); 
for(i=0; hex[i]!='\0'; ++i, --length) 
{ 
    if(hex[i]>='0' && hex[i]<='9') 
     decimal+=(hex[i]-'0')*pow(16,length-1); 
    if(hex[i]>='A' && hex[i]<='F') 
     decimal+=(hex[i]-55)*pow(16,length-1); 
    if(hex[i]>='a' && hex[i]<='f') 
     decimal+=(hex[i]-87)*pow(16,length-1); 
} 

//At this point, variable decimal contains the hexadecimal number in decimal format. 

    i=1; 
    while (decimal!=0) 
    { 
     binary+=(decimal%2)*i; 
     decimal/=2; 
     i*=10; 
    } 
    return binary; 
} 

Answer 1

您需要使用一个数组来存储二进制数，作为一个int变量将不能够存储大量数，int范围（通常）从−32767到+32767。 C data types.

例如：

---

#include <stdio.h> 
#include <math.h> 

int hex_binary(char hex[], int binary_number[]); 

int main() 
{ 
    char hex[20]; 
    int binary_number[100]; 
    int j,k; 

    printf("Enter hexadecimal number: "); 
    scanf("%19s",hex); 
    j = hex_binary(hex,binary_number); 

    printf("Binary number is: "); 
    for(k=j-1;k>=0;k--) 
     printf("%d",binary_number[k]); 
    printf("\n"); 

    return 0; 
} 

//Function to convert hexadecimal to binary. 

int hex_binary(char hex[], int binary_number[]) 
{ 
    int i, j=0, length, decimal=0; 
    for(length=0; hex[length]!='\0'; ++length); 
    for(i=0; hex[i]!='\0'; ++i, --length) 
    { 
     if(hex[i]>='0' && hex[i]<='9') 
      decimal+=(hex[i]-'0')*pow(16,length-1); 
     if(hex[i]>='A' && hex[i]<='F') 
      decimal+=(hex[i]-55)*pow(16,length-1); 
     if(hex[i]>='a' && hex[i]<='f') 
      decimal+=(hex[i]-87)*pow(16,length-1); 
    } 
    //At this point, variable decimal contains the hexadecimal number in decimal format. 
    while (decimal!=0) 
    { 
     binary_number[j++] = decimal%2; 
     decimal/=2; 
    } 
    return j; 
} 

输出

Enter hexadecimal number: 111 
Binary number is: 100010001 

Enter hexadecimal number: 1111 
Binary number is: 1000100010001 

Enter hexadecimal number: 11111 
Binary number is: 10001000100010001 

Enter hexadecimal number: 987634 
Binary number is: 100110000111011000110100