如何计算C中矩阵的一行中的元素数量

Question

使用整数数组，Id首先要询问用户在数组中想要的行和列的数量（我称它们为x和y）（我知道如何执行此步骤）。 重要的是，尽管用户输入了将存储在矩阵中的值，但将从一行输入中读取一行，从第二行输入中读取第二行，依此类推。 因此，输入的一行等于1行。

因此，如果他定义了x列数，则应该在第一行中全部输入x个数字。 我如何检查是否确实在该行输入了x个数字？ 如果输入的更多或更少，我将打印一条错误消息。 是否有某种命令可以检查1行的大小，以便可以根据用户定义的x进行检查？

到目前为止，我编写的代码涉及简单的步骤，但是对于如何执行此检查，以确认用户输入的输入量与他最初定义的输入量相同，我几乎一无所知。

非常感谢！

Answer 1

由于您想一次读取一行，因此应该使用fgets来做到这一点。 然后，您可以使用strtok用空格/制表符分隔行，并尝试将每个值转换为int。 如果在行填充之前数字用完了，或者在行填充之后仍然有更多数字，则可以提醒用户。

这是一个简单的示例：

void read_array(int **array, int rows, int cols)
{
    char line[100];
    int i,count;
    char *p;

    for (i=0;i<rows;i++) {
        printf("enter row %d: "), i-1);
        fgets(line, sizeof(line), stdin);
        count = 0;
        p = strtok(line, " \t");
        while (p && (count < cols)) {
            matrix[i][count++] = atoi(p);
            p = strtok(NULL, " \t");
        }
        if (count < cols) {
            printf("too few values\n");
            i--;
            continue;
        } else if (p) {
            printf("too many values\n");
            i--;
            continue;
        }
    }
}

Answer 2

这是另一种解决方案，尽管对于您的特定要求，其他答案比我的要好。

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

int main(void)
{
    int number_of_rows;
    int number_of_columns;

    printf("Enter the number of rows: ");
    scanf("%d", &number_of_rows);
    printf("Ok, enter the number of columns: ");
    scanf("%d", &number_of_columns);

    int matrix[number_of_rows][number_of_columns];

    int i;
    int j;

    for(i = 0; i < number_of_rows; ++i) {
        printf("This is %d row.\n", i+1);
        for(j = 0; j < number_of_columns; ++j) {
            scanf("%d", &matrix[i][j]);
        }
    }

    /* Print's the 2D array. */
    for(i = 0; i < number_of_rows; ++i) {
        printf("\n");
        for(j = 0; j < number_of_columns; ++j) {
            printf("%d   ", matrix[i][j]);
        }
        printf("\n");
    }

    return 0;
}

这是在向用户询问用户希望矩阵（2D数组）具有的行数和列数，然后根据这些数字声明一个2D数组。 然后，它遍历每一行，在遍历每一列之前，将控制台的行号打印到控制台。 初始化2D数组后，它将以绘制矩阵的方式打印2D数组。 我希望这有帮助。

Answer 3

这使用一个函数获取一个范围内的整数，并使用另一个函数获取一系列整数。 strtol用于从使用fgets获得的输入中解析值。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <errno.h>
#include <limits.h>

//inputs
// char *line : pointer to text to be parsed
// char **next : pointer to pointer to allow modification of caller's pointer
// char *term : pointer to characters to be considered terminators
// int *value : pointer to int to allow modification of caller's int
// int min : minimum value of range
// int max : maximum value of range
// returns : 0 failure or 1 success
int get_int_range ( char *line, char **next, char *term, int *value, int min, int max)
{
    long int input = 0;
    char *end = NULL;//will point to end of parsed value

    errno = 0;
    input = strtol ( line, &end, 10);//get the integer from the line. end will point to the end of the parsed value
    if ( end == line) {// nothing was parsed. no digits
        printf ( "input MUST be a number\n");
        return 0;// return failure
    }
    // *end is the character that end points to
    if ( *end != '\0' && ( strchr ( term, *end) == NULL)) {// is *end '\0' or is *end in the set of term characters
        printf ( "problem with input: [%s] \n", line);
        return 0;
    }
    if ( ( errno == ERANGE && ( input == LONG_MAX || input == LONG_MIN))
    || ( errno != 0 && input == 0)){// parsing error from strtol
        perror ( "input");
        return 0;
    }
    if ( input < min || input > max) {// parsed value is outside of range
        printf ( "input out of range %d to %d\n", min, max);
        return 0;
    }

    if ( next != NULL) {// if next is NULL, caller did not want pointer to end of parsed value
        *next = end;// *next allows modification to caller's pointer
    }
    *value = input;// *value allows modification to callers int
    return 1;// success
}

//inputs
// int cols : needed to pass in the variable length array
// int dest[][cols] : variable length array to be modified with parsed values
// int inputrow : row of the variable length array to be modified
// int min : mininum value of range
// int max : maximum value of range
// char *line : pointer to text to be parsed
// char *term : pointer to characters to be considered terminators
// returns : 0 failure or 1 success
int get_int_series ( int cols, int dest[][cols], int inputrow, int min, int max, char *line, char *term)
{
    char *end = NULL;// will point to end of parsed value
    char *each = NULL;// will point to start of parsed value
    int valid = 0;
    int input = 0;
    int count = 0;
    int temp[cols];// temporary storage for parsed values

    each = line;// each points to start of line
    do {
        valid = get_int_range ( each, &end, term, &input, INT_MIN, INT_MAX);// call to parse one value
        if ( !valid) {// failure
            printf ( "input MUST be a number\n");
            return 0;
        }
        if ( valid) {// success
            temp[count] = input;// save parsed value in temporary array
            count++;
            if ( count > cols) {
                printf ( "too many integers. %d entered. only enter %d\n", count, cols);
                return 0;
            }
        }
        // *end is the character that end points to
        while ( *end && strchr ( term, *end)) {// if *end not '\0' and *end is in the set of term characters
            end++;// yes. advance end one position
        }
        each = end;// set the end pointer to be the new starting pointer
    } while ( end && *end);// keep looping the above while end is not NULL and *end is not '\0' 

    if ( count < cols) {
        printf ( "too few integers. need %d entered. only entered %d\n", cols, count);
        return 0;
    }
    while ( count) {
        count--;
        dest[inputrow][count] = temp[count];// set the values in the callers array to the values in the temporary array
    }
    return 1;// success
}

int main( int argc, char *argv[])
{
    char line[900] = {'\0'};
    int valid = 0;
    int rows = 0;
    int cols = 0;
    int eachrow = 0;
    int eachcol = 0;

    do {
        printf ( "Enter number of array rows or quit\n");
        fgets ( line, sizeof ( line), stdin);//read a line
        if ( strcmp ( line, "quit\n") == 0) {
            return 1;// if quit is entered, exit the program
        }
        valid = get_int_range ( line, NULL, "\n", &rows, 0, INT_MAX);// call to parse a value
    } while ( !valid);// on failure, keep looping the above

    do {
        printf ( "Enter number of array columns or quit\n");
        fgets ( line, sizeof ( line), stdin);//read a line
        if ( strcmp ( line, "quit\n") == 0) {
            return 1;
        }
        valid = get_int_range ( line, NULL, "\n", &cols, 0, INT_MAX);
    } while ( !valid);

    //once the size is obtained, the variable length array can be declared
    int array[rows][cols];

    for(eachrow = 0; eachrow < rows; eachrow++) {// loop through the rows
        do {
            printf ( "Enter %d integers for row %d or quit. Ex 1 2 3\n", cols, eachrow + 1);
            fgets ( line, sizeof ( line), stdin);//read a line
            if ( strcmp ( line, "quit\n") == 0) {
                return 1;
            }
            valid = get_int_series ( cols, array, eachrow, INT_MIN, INT_MAX, line, " \n");// call to obtain a series of values
        } while ( !valid);// on failure, keep looping the above
    }
    // show the values in the variable length array
    for(eachrow = 0; eachrow < rows; eachrow++) {
        for(eachcol = 0; eachcol < cols; eachcol++) {
            printf("[%d] ", array[eachrow][eachcol]);
        }
        printf("\n");
    }
    printf("\nDone\n");

    return 0;
}