我如何用C將二進制結果分成4組。代碼的目的是將十進制數字轉換為二進制表示法

Question

例如， 233的二進制轉換是11101001，並且想要將其轉換為1110 | 1001之類的東西

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

int main()
{
    int i, j, r, decimal, binary[20];

    while(1)
    {
        printf("\n\nPLEASE ENTER ANY DECIMAL NUMBER: ");
        scanf("%d", &decimal);
        int n = decimal;

        i = 0;

        while (decimal > 0)
        {
            r = decimal % 2;
            binary[i] = r;
            decimal = decimal / 2;
            i++;
        }
        printf("\n BINARY EQUIVALENT OF %d is; ", n);

        for (j = i - 1; j >= 0; j--)
        {
            printf("%d", binary[j]);
        }
    }
    return 0;
}

我敢打賭，對於某些人來說，這是一個菜鳥問題，但我可以使用您的幫助，謝謝

Answer 1

有趣的是，我昨天剛剛這樣做。 這可能不是執行此操作的最佳方法，但它可以工作。

char* hex_to_bin (unsigned char c) {
    static char bin[10];
    strcpy (bin, "");
    char* mappings[] = {
        "0000",
        "0001",
        "0010",
        "0011",
        "0100",
        "0101",
        "0110",
        "0111",
        "1000",
        "1001",
        "1010",
        "1011",
        "1100",
        "1101",
        "1110",
        "1111"
    };

    sprintf (bin, "%s %s", mappings [(c/16) % 16], mappings[c % 16]);

    return bin;
}

void print_bin (void * ptr, size_t size) {
    int i;
    for (i = size - 1; i >= 0 ; i--) {
        unsigned char c = ((char*)ptr)[i];
        printf ("%s ", hex_to_bin(c));
    }
}

使用方法：

int k = 431;
print_bin (&k, sizeof (int)); // Of course you can put any pointer here

它應該打印此：

0000 0000 0000 0000 0000 0001 1010 1111

注意，在for我從size迭代到0。這特定於Little Endian處理器。 如果您有Big Endian處理器，則必須從0擴展到size

Answer 2

例如，您可以在“ j”的第二個外觀中添加條件，並在打印4位數字后添加|。

    for (int j=i-1; j>=0; j--) {
        printf( "%d",binary[j]);
        if(j%4==0 && j!=0)
            printf("|");   
    }

Answer 3

如果只想打印它，建議使用按位操作數和位移 。 對C和按位操作數中的移位（如果您當然還不知道）進行一些研究，然后查看下面的代碼和說明。 這是一個更簡化的代碼，並且性能也更高，並且您將通過變量聲明節省內存。

很少解釋： 按位操作數比較一個或兩個變量的每個單個位。 通過位移位，可以將變量的位向所需次數的方向移位 。

參見下面的代碼：

#include <stdio.h>

void main() {

    int i, decimal;
    //The mask to compare just one bit
    int mask=0b10000000;

    printf("\n\nPLEASE ENTER ANY DECIMAL NUMBER: ");
    scanf("%d", &decimal);

    //Loopping 8 bits only
    for(i=0;i<8;i++) {
        /**
        * Here you are doing the "magic", see how:
        * You're using AND bitwise operand to compare just one bit at time starting for the MSB as you can see in first mask value
        * So, first comparison with AND is you're value, for example 254, with mask shifted of 'i' bits, i value is 0 here
        * The first comparison so is: (255)11111110   -> YOUR INT VALUE
        *                           (AND)& 10000000   -> MASK
        *                          Result: 10000000   -> THE RESULT OF: decimal&mask
        * But this result in decimal is 128 and you want to show just 1 or 0, being 1 for everything that is not and 0 and 0 only for 0 results
        * Here I suggest a technique, use the NOT logical operand to invert the value logically and the do it again, as bellow:
        * !10000000 result in 0 and !0 result in 1 or 00000001, this way you can show if the bit is HIGH or DOWN.
        * Keep doing it for the other bits, let see more 2:
        * -----------------------------------------------------------------------------------------------
        * Now i = 1 (second time in for loop), so mask is shifted 1 bit to right and mask is 01000000
        * Making the same comparison (255)11111110   -> YOUR INT VALUE
        *                          (AND)& 01000000   -> MASK
        *                         Result: 01000000   -> THE RESULT OF: decimal&mask
        * When you invert the result twice the result is 1 again.
        * ------------------------------------------------------------------------------------------------
        * Now i = 7 (last (seventh) time in for loop), so mask is shifted 7 bits to right and mask is 00000001
        * Making the same comparison (255)11111110   -> YOUR INT VALUE
        *                          (AND)& 00000001   -> MASK
        *                         Result: 00000000   -> THE RESULT OF: decimal&mask
        * When you invert the result twice the result is now 0. Because 0&1 is 0, when inverted it is 1 and inverting again it turns 0.
        *
        */
        printf("%d", !(!(decimal&mask>>i)));
        if(i==3) printf(" "); //If it is on the fourth bit will print a space
    }
}

Answer 4

您可以使用掩碼，按位運算符可將<<和AND＆移位。 如果逐步調試器中的代碼，您將看到它是如何工作的。

void displayBits(unsigned);

int main(int args, char *argsv[]){

   unsigned x;
   printf("Enter an unsigned integer %u\n", x );
   scanf("%u",&x);
   displayBits(x);

    return 0;
}

void displayBits(unsigned value){
    const int SHIFT = 8*sizeof(unsigned)-1;
    const unsigned MASK = 1 << SHIFT;

    printf("%u = ", value);

    unsigned c;
    for(c=1; c<=SHIFT + 1; c++){

        if(value & MASK){
            printf("%d", 1);

        }
        else
        {
            printf("%d",0);

        }
        if(c%8==0)
        printf("|");
        value <<=1;
    }
}