c语言的 浮点数计算程序纠错

Question

我想写一个c语言的 浮点数计算程序，目前是半精度（16位）计算出了问题，单精度（32位）计算正常。目前程序输入
1.0
0.0
16
16
输出的结果是0.00000。
如果是输入1.0 0.0 32 32则输出结果为1.00000这个结果是正确的。请问如何修改这个程序

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

// 把浮点数转换为二进制表示
char * valueToMachine(float num, int length){
    unsigned int n = *((unsigned int*)&num);
    char * binary = (char*)malloc(length + 1);
    if(binary == NULL){
        printf("Memory allocation failed\n");
        exit(1);
    }
    binary[length] = '\0';
    if(length == 32){
        for(int i = 31; i >= 0; i--){
            binary[i] = (n & 1) ? '1' : '0';
            n >>= 1;
        }
    }else if(length == 16){
        // handle the sign bit, exponent bits and fraction bits separately
        binary[0] = (n >> 15) ? '1' : '0'; // sign bit
        int exponent = (n >> 10) & 0x1F; // exponent bits
        int fraction = n & 0x3FF; // fraction bits
        for(int i = 1; i <= 5; i++){ 
            binary[i] = ((exponent >> (5 - i)) & 1) ? '1' : '0';
        }
        for(int i = 6; i <= 15; i++){ 
            binary[i] = ((fraction >> (15 - i)) & 1) ? '1' : '0';
        }
    }
    return binary;
}

// 把二进制表示转换回浮点数
float machineToValue(char * binary, int length){
    unsigned int n = 0;
    if(length == 32){
        for(int i = 0; i < length; i++){
            n = (n << 1) + (binary[i] == '1' ? 1 : 0);
        }
    } else if(length == 16){
        unsigned int sign = binary[0] == '1' ? 0x8000 : 0; // sign bit
        unsigned int exponent = 0;
        for(int i = 1; i <= 5; i++){ 
            exponent = (exponent << 1) + (binary[i] == '1' ? 1 : 0);
        }
        unsigned int fraction = 0;
        for(int i = 6; i <= 15; i++){ 
            fraction = (fraction << 1) + (binary[i] == '1' ? 1 : 0);
        }
        // Convert back to IEEE 754 single-precision format
        if(exponent == 0 && fraction == 0){
            return (sign == 0) ? 0.0 : -0.0; // Handle the special case of +0 and -0
        } else if(exponent == 0x1F){ //处理特殊值
            return fraction == 0 ? ((sign == 0) ? INFINITY : -INFINITY) : NAN;
        } else{
            n = sign << 15 | ((exponent - 15 + 127) << 10) | fraction; // Add bias to the exponent
        }
    }
    float num;
    memcpy(&num, &n, sizeof(num));
    return num;
}


// 计算两个二进制表示的浮点数的和
char * addfloat(char * float1, char * float2, int length){
    float num1 = machineToValue(float1, length);
    float num2 = machineToValue(float2, length);
    float sum = num1 + num2;
    return valueToMachine(sum, 32); // the result is always in single-precision floating-point format
}

// 计算两个浮点数的和
float computing(float float1, float float2, int n1, int n2){
    char * binary1 = valueToMachine(float1, n1);
    char * binary2 = valueToMachine(float2, n2);
    char * binary_sum = addfloat(binary1, binary2, n1 > n2 ? n1 : n2);
    float sum = machineToValue(binary_sum, 32);
    free(binary1);
    free(binary2);
    free(binary_sum);
    return sum;
}

int main(){
    float float1, float2;
    int n1, n2;
    int result = scanf("%f%f%d%d", &float1, &float2, &n1, &n2);
    if(result != 4 || (n1 != 16 && n1 != 32) || (n2 != 16 && n2 != 32)){
        printf("Invalid input\n");
        return 1;
    }
    float sum = computing(float1, float2, n1, n2);
    printf("%f\n", sum);
    return 0;
}

Answer 1

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
 
// 把浮点数转换为二进制表示
char * valueToMachine(float num, int length){
    unsigned int n = *((unsigned int*)&num);
    char * binary = (char*)malloc(length + 1);
    if(binary == NULL){
        printf("Memory allocation failed\n");
        exit(1);
    }
    binary[length] = '\0';
    if(length == 32){
        for(int i = 31; i >= 0; i--){
            binary[i] = (n & 1) ? '1' : '0';
            n >>= 1;
        }
    }else if(length == 16){
        // handle the sign bit, exponent bits and fraction bits separately
        binary[0] = (n >> 31) ? '1' : '0'; // sign bit
        int exponent = ((n >> 23) & 0xFF) - 127 + 15; // exponent bits
        if (exponent < 0) exponent = 0;
        if (exponent > 0x1F) exponent = 0x1F;
        int fraction = (n >> 13) & 0x3FF; // fraction bits
        for(int i = 1; i <= 5; i++){ 
            binary[i] = ((exponent >> (5 - i)) & 1) ? '1' : '0';
        }
        for(int i = 6; i <= 15; i++){ 
            binary[i] = ((fraction >> (15 - i)) & 1) ? '1' : '0';
        }
    }
    return binary;
}
 
// 把二进制表示转换回浮点数
float machineToValue(char * binary, int length){
    unsigned int n = 0;
    if(length == 32){
        for(int i = 0; i < length; i++){
            n = (n << 1) + (binary[i] == '1' ? 1 : 0);
        }
    } else if(length == 16){
        unsigned int sign = binary[0] == '1' ? 0x80000000 : 0; // sign bit
        unsigned int exponent = 0;
        for(int i = 1; i <= 5; i++){ 
            exponent = (exponent << 1) + (binary[i] == '1' ? 1 : 0);
        }
        unsigned int fraction = 0;
        for(int i = 6; i <= 15; i++){ 
            fraction = (fraction << 1) + (binary[i] == '1' ? 1 : 0);
        }
        // Convert back to IEEE 754 single-precision format
        n = sign | ((exponent == 0 ? 0 : exponent - 15 + 127) << 23) | (fraction << 13);
    }
    float num;
    memcpy(&num, &n, sizeof(num));
    return num;
}
 
// 计算两个二进制表示的浮点数的和
char * addfloat(char * float1, char * float2, int length){
    float num1 = machineToValue(float1, length);
    float num2 = machineToValue(float2, length);
    float sum = num1 + num2;
    return valueToMachine(sum, 32); // the result is always in single-precision floating-point format
}
 
// 计算两个浮点数的和
float computing(float float1, float float2, int n1, int n2){
    char * binary1 = valueToMachine(float1, n1);
    char * binary2 = valueToMachine(float2, n2);
    char * binary_sum = addfloat(binary1, binary2, n1 > n2 ? n1 : n2);
    float sum = machineToValue(binary_sum, 32);
    free(binary1);
    free(binary2);
    free(binary_sum);
    return sum;
}
 
int main(){
    float float1, float2;
    int n1, n2;
    int result = scanf("%f%f%d%d", &float1, &float2, &n1, &n2);
    if(result != 4 || (n1 != 16 && n1 != 32) || (n2 != 16 && n2 != 32)){
        printf("Invalid input\n");
        return 1;
    }
    float sum = computing(float1, float2, n1, n2);
    printf("%f\n", sum);
    return 0;
}