Question

程序从文本文件中输入一个字符串，如：

2 1.0 2.0 3.0 4.0

其中第一个数字是方形矩阵的维数，其他数字是矩阵的元素（以列主要形式存储）。每个数字都由其他人用“空格”字符分隔该程序在一列上添加所有数字，并将每种结果相乘。 I.E.使用此字符串，结果将是：21.0 问题是：使用此输入，程序的输出将是：

输入字符串是2 1.0 2.0 3.0 4.0




从字符串中提取char类型标记1.0

从字符串中提取char类型标记2.0

从字符串中提取char类型标记3.0

从字符串中提取char类型标记4.0

将char类型标记1.0转换为float类型

将char类型令牌2.0转换为float类型

将char类型令牌2.0转换为float类型

将char类型标记3.0转换为浮点类型


浮动类型的打印矩阵：

1.000000

2.000000

2.000000

3.000000

最终结果是15.000000

相反它应该是：输入字符串是2 1.0 2.0 3.0 4.0

从字符串中提取char类型标记1.0

从字符串中提取char类型标记2.0

从字符串中提取char类型标记3.0

从字符串中提取char类型标记4.0

将char类型标记1.0转换为float类型将转换后的char类型标记2.0转换为

float类型将char类型标记2.0转换为float类型转换

char type token 3.0 into float type


浮动类型的打印矩阵：

1.000000

2.000000

3.000000

4.000000

最终结果为24.000000

这里是代码

#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>




/* void calculus(char string[], int matrixDimension)
 * Executes the following procedure:
 * 1) it extracts from the string, various char type tokens
 * 2) converts these char type tokens into float types and stores it into "squareMatrix" bidimensional vector
 * 3) does the calculus
 */
void calculus(char string[]);





int main()
{
    FILE* fileToReadFd;
    int nRead;
    char string[500] = {0};
    const char pathNameRead[] = "/home/caterpillar/canc/matrice2.txt";
    if((fileToReadFd = fopen(pathNameRead, "r")) == NULL)
    {
        printf("Ho provato ad aprire %s\n", pathNameRead);
        printf("errore nell'aprire il file\n" "%s\n", strerror(errno));
    }
    nRead=fread(&string[0],sizeof(char),100,fileToReadFd);


    printf("Input string is %s\n", &string[0]);
    calculus(string);
    fclose(fileToReadFd);
    return 0;

}


void calculus(char string[])
{
    int matrixDimension = atoi(&string[0]);

    float finalResult = 1;

    // float type square matrix to be filled
    float squareMatrix[matrixDimension][matrixDimension];

    // stores the result of every column addition
    float columnAddition[matrixDimension];

    /*
     * stores tokens from the string
     * I.E.:
     * token[0] contains "1.0"
     * token[1] contains "2.0"
     * token[2] contains "3.0"
     * token[3] contains "4.0"
     */
    char tokens[matrixDimension * matrixDimension][8];

    /*
     * zero initialize columnAddition vector
     */
    for(int i = 0; i < matrixDimension; i++)
    {
        columnAddition[i] = 0;
    }

    /*
     *  First strtok is necessary to be left alone since it takes away
     *  the first token that is not usefull ( it is the matrix dimension)
     */
    strtok(&string[0], " ");
    for(int i = 0; i < (matrixDimension * matrixDimension); i++)
    {
        strcpy(&tokens[i][0], strtok(NULL, " "));
        printf("Extracted char type token %s from the string\n", &tokens[i][0]);
    }

    for(int i = 0; i < matrixDimension; i++)
    {
        for(int j = 0; j < matrixDimension; j++)
        {
            squareMatrix[i][j] = atof(&tokens[i+j][0]);
            printf("Converted char type token %s into float type\n", &tokens[i+j][0]);
        }
    }
    printf("\nPrinting matrix of float types:\n");
    for(int i = 0; i < matrixDimension; i++)
    {
        for(int j = 0; j < matrixDimension; j++)
        {
            printf("%f\n", squareMatrix[i][j]);
        }
    }
    // does calculus
    for(int j = 0; j < matrixDimension; j++)
    {
        for(int i = 0; i < matrixDimension; i++)
        {

            columnAddition[j] = columnAddition[j] + squareMatrix[i][j];
        }
    }
    for(int i = 0; i < matrixDimension; i++)
    {
        finalResult = finalResult * columnAddition[i];
    }
    printf("Final result is %f\n", finalResult);
}

Answer 1

我认为这是问题：

squareMatrix[i][j] = atof(&tokens[i+j][0]);
printf("Converted char type token %s into float type\n", &tokens[i+j][0]);

因为这将为（i = 0，j = 1）和（i = 1，j = 0）引用相同的标记，因为1 + 0 = 1和0 + 1 = 1

应该是

squareMatrix[i][j] = atof(&tokens[i*matrixDimension+j][0]);
printf("Converted char type token %s into float type\n", &tokens[i*matrixDimension+j][0]);

矩阵运算，从char到float的错误转换

1 个答案: