Question

嘿，我得到了这个错误，我尝试了10个解决方案，并且都有效。我想加载2个矩阵，每个矩阵来自它自己的txt文件，然后将它们相乘。我无法编译LNK1120和LNK2019错误的casue。这是我的代码：

int main(int argc, char *argv[])
{


FILE *macierz1, *macierz2, *fw;
char *line = malloc(1000);
int count = 0;
macierz1 = fopen("macierz1.txt", "r");
    if (macierz1 == NULL) {``
    printf("nie można otworzyć", argv[1]);
    exit(1);
}


macierz2 = fopen("macierz2.txt", "r");
if (macierz2 == NULL) {
    printf("nie można otworzyć", argv[2]);
    exit(1);
}

double *data = (double*)malloc(1000 * sizeof(double));
if (data == NULL)
{
    printf("błąd lokowania pamięci");
    return EXIT_FAILURE;
}
getline(&line, &count, macierz1);
int read = -1, cur = 0, columCount1 = 0;
while (sscanf(line + cur, "%lf%n", &data[columCount1], &read) == 1)
{
    cur += read; columCount1++;
}

int rowCount1 = 1;
while (getline(&line, &count, macierz1) != -1) { rowCount1++; }
printf("%d\n", columCount1);
printf("%d\n", rowCount1);

getline(&line, &count, macierz2);
read = -1, cur = 0;
int columCount2 = 0;
while (sscanf(line + cur, "%lf%n", &data[columCount2], &read) == 1)
{
    cur += read; columCount2++;
}

int rowCount2 = 1;
while (getline(&line, &count, macierz2) != -1) { rowCount2++; }
printf("%d\n", columCount2);
printf("%d\n", rowCount2);
int i = 0;
int j = 0;

int **mat1 = (int **)malloc(rowCount1 * sizeof(int*));
for (i = 0; i < rowCount1; i++)
    mat1[i] = (int *)malloc(columCount1 * sizeof(int));

fseek(macierz1, 0, SEEK_SET);

for (i = 0; i < rowCount1; i++)
{
    for (j = 0; j < columCount1; j++)
        fscanf(macierz1, "%d", &mat1[i][j]);
}

i = 0;
j = 0;

printf("\n\n");
//print matrix 1
for (i = 0; i < rowCount1; i++)
{
    for (j = 0; j < columCount1; j++)
        printf("%d", mat1[i][j]);

    printf("\n");
}

i = 0;
j = 0;
int **mat2 = (int **)malloc(rowCount2 * sizeof(int*));
for (i = 0; i < rowCount2; i++)
    mat2[i] = (int *)malloc(columCount2 * sizeof(int));

fseek(macierz2, 0, SEEK_SET);

for (i = 0; i < rowCount2; i++)
{
    for (j = 0; j < columCount2; j++)
        fscanf(macierz2, "%d", &mat2[i][j]);
}

i = 0;
j = 0;

printf("\n\n");
//print matrix 2
for (i = 0; i < rowCount2; i++)
{
    for (j = 0; j < columCount2; j++)
        printf("%d", mat2[i][j]);

    printf("\n");
}

i = 0;

int **mat3 = (int **)malloc(rowCount1 * sizeof(int*));
for (i = 0; i < rowCount1; i++)
    mat3[i] = (int *)malloc(columCount2 * sizeof(int));
i = 0;
j = 0;
int k = 0;
int sum = 0;

if (columCount1 != rowCount2)
{
    puts("The number of columns in Matrix 1 is not same as the number of rows in Matrix 2");
    exit(1);
}

//multiplication of two matrices
for (i = 0; i<rowCount1; i++)
{
    for (j = 0; j<columCount2; j++)
    {
        mat3[i][j] = 0;
        for (k = 0; k<columCount1; k++)
        {
            mat3[i][j] = mat3[i][j] + mat1[i][k] * mat2[k][j];
        }
    }
}


//print multiplication result
printf("\n\nResult = \n\n");

for (i = 0; i < rowCount1; i++)
{
    for (j = 0; j < columCount2; j++)
        printf("%d", mat3[i][j]);

    printf("\n");
}
for (i = 0; i< rowCount1; i++)
    free(mat1[i]);
free(mat1);

for (i = 0; i< rowCount2; i++)
    free(mat2[i]);
free(mat2);

for (i = 0; i< rowCount1; i++)
    free(mat3[i]);
free(mat3);

free(data);

return 0;
}

Answer 1

由于您使用的是Visual C ++，因此C函数没有getline。 Visual C ++附带的C编译器遵循C89规范。

使用fgets从文件中检索数据。

如果您真的想在Visual C ++（真正的C）程序中使用getline，可以尝试this code from GNU来实现该功能。

Answer 2

有一个流浪的＆＃39;`＆＃39;在程序结尾处发布的程序中：

if (macierz1 == NULL) {``

和

getline

功能未定义。

添加所需的包括：

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

无法编译矩阵乘法（C）

2 个答案: