我想获得数组的每一行和每列的总和，并在C中逐一打印。我完成了一半的工作，但我不能再进步了。那是我的代码：
int n,m;
void sum_row_column(int array[n][m],int r,int c,int i,int j)
{
int sum_of_column1=0;
int sum_of_row1;
for(j=0,i=0;i<r;i++)
{
sum_of_column1=sum_of_column1+array[i][j];
if(i==r-1)
{
printf("\nSum of %d.column: %d\n",j+1,sum_of_column1);
j=j+1;
}
}
for(i=0,j=0;j<r;j++)
{
sum_of_row1=sum_of_row1+array[i][j];
if(j=c-1)
{
printf("\nSum of %d.row: %d\n",j+1,sum_of_row1);
i=i+1;
}
}
正如您所看到的，它只打印第一行的总和和列的总和。我想如果我添加
i=0;
后
printf("\nSum of %d.column: %d\n",j+1,sum_of_column1);
j=j+1
行，它可以移动第二列并将其相加。但是当我添加
i=0;
程序进入无限循环而没有任何错误。 这段代码在标题中，但不要担心其余的代码没有任何错误。
答案 0 :(得分：1)
你有：
int n,m;
void sum_row_column(int array[n][m],int r,int c,int i,int j)
{
虽然这是编译的，但它是定义不明确的代码，如果未正确设置全局变量n和m，则会不必要地失败。不需要那种级别的耦合，如果使用该级别的耦合，变量r和c是多余的。你应该使用：
void print_sum_rows_cols(int r, int c, int array[r][c])
{
这允许代码用于使用VLA（可变长度数组）符号打印任何形状的矩阵。变量n和m应该是不必要的;参数i和j似乎也毫无意义，因为您忽略了传递给函数的值，在循环开始时将它们都设置为0。
对列进行求和很简单：
for (int col = 0; col < c; col++)
{
int colsum = 0;
for (int row = 0; row < r; row++)
colsum += array[row][col];
printf("Sum for col %d = %d\n", col, colsum);
}
对行进行求和同样是直截了当的：
for (int row = 0; row < r; row++)
{
int rowsum = 0;
for (int col = 0; col < c; col++)
rowsum += array[row][col];
printf("Sum for row %d = %d\n", row, rowsum);
}
组装MCVE（Minimal, Complete, Verifiable Example）：
#include <stdio.h>
static void print_sum_rows_cols(int r, int c, int array[r][c])
{
for (int col = 0; col < c; col++)
{
int colsum = 0;
for (int row = 0; row < r; row++)
colsum += array[row][col];
printf("Sum for col %d = %d\n", col, colsum);
}
for (int row = 0; row < r; row++)
{
int rowsum = 0;
for (int col = 0; col < c; col++)
rowsum += array[row][col];
printf("Sum for row %d = %d\n", row, rowsum);
}
}
static void dump_matrix(const char *tag, int rows, int cols, int matrix[rows][cols])
{
printf("%s (%dx%d):\n", tag, rows, cols);
for (int r = 0; r < rows; r++)
{
for (int c = 0; c < cols; c++)
printf("%3d", matrix[r][c]);
putchar('\n');
}
}
int main(void)
{
int a1[3][4] =
{
{ 68, 78, 50, 46, },
{ 64, 12, 47, 1, },
{ 86, 10, 84, 62, },
};
int a2[5][3] =
{
{ 4, 30, 19, },
{ 79, 58, 20, },
{ 95, 12, 24, },
{ 20, 37, 72, },
{ 17, 0, 53, },
};
dump_matrix("A1", 3, 4, a1);
print_sum_rows_cols(3, 4, a1);
dump_matrix("A2", 5, 3, a2);
print_sum_rows_cols(5, 3, a2);
return 0;
}
（重命名）print_sum_rows_cols()函数中的代码维护问题的变量名称，尽管它们不是我在自己的代码中使用的名称。 （对于那些，请参阅“另一个答案的分析”部分。）
输出：
A1 (3x4):
68 78 50 46
64 12 47 1
86 10 84 62
Sum for col 0 = 218
Sum for col 1 = 100
Sum for col 2 = 181
Sum for col 3 = 109
Sum for row 0 = 242
Sum for row 1 = 124
Sum for row 2 = 242
A2 (5x3):
4 30 19
79 58 20
95 12 24
20 37 72
17 0 53
Sum for col 0 = 215
Sum for col 1 = 137
Sum for col 2 = 188
Sum for row 0 = 53
Sum for row 1 = 157
Sum for row 2 = 131
Sum for row 3 = 129
Sum for row 4 = 70
请注意，这会打印输入，以便可以根据输入手动检查输出。这是一种有用的调试技术。
在answer（修订版1）中，Jozeph显示了可用的代码。当以正统格式缩进时，代码为：
int n; int m; void sum_row_column(int array[n][m], int r, int c, int i, int j) { int sumOfColumn = 0; int sumOfRow = 0; printf("\n"); for (i = 0; i < r; i++) { sumOfRow = 0; for (j = 0; j < c; j++) { sumOfRow += array[i][j]; } printf("Sum of %d.row=%d\n", i + 1, sumOfRow); } printf("\n"); for (j = 0; j < r; j++) { sumOfColumn = 0; for (i = 0; i < c; i++) { sumOfColumn += array[i][j]; } printf("Sum of %d.column=%d\n", j + 1, sumOfColumn); } }
此代码存在多个问题，一些问题已经确定，但在这里以不同的形式重复：
n和m中的值标识。r和c标识。i和j中的值未使用;它们只是定义两个局部循环变量的一种特殊方式。assert(n == r && m == c);，第一对嵌套循环（对行进行求和）就可以了。assert(n == r && m == c && n == m):（方阵），第二对嵌套循环（对列进行求和）是可以的，但它主要是偶然而不是设计。r x c，但函数假定数组的大小为m x n。这是我原来答案的改编，包括建议的答案（如上所示），并添加了一些调试（只读）打印。它还包括我原始建议函数的一个小变体 - 变量被重命名以适应我的偏见（循环变量的短名称，函数参数的较长名称等），以及调试打印代码。该代码还在列总和之前打印行总和以匹配建议的答案序列，并且它标识从第1行开始，第1列而不是第0行第0列的行。
我的答案（bogus_sum_rows_cols()）中还有一个'虚假'变体，它使用n和m变量来调整数组的大小;当n和m与r和c不同步时，它会生成错误的答案。如果bogus_sum_row_column()和r中的数组大小与c和{n中的数组大小匹配，那么建议答案（m）的“虚假”变体也会至少正确地对列进行求和。 {1}}。
/* SO 47719-9719 - Abuse of VLA definitions (and notations) */
#include <stdio.h>
static int debug = 0;
// Original answer (renamed variables, reordered results, and print row/col + 1)
static void print_sum_rows_cols(int rows, int cols, int array[rows][cols])
{
for (int r = 0; r < rows; r++)
{
int rowsum = 0;
for (int c = 0; c < cols; c++)
{
if (debug)
printf(" a[%d][%d] = %d", r, c, array[r][c]);
rowsum += array[r][c];
}
if (debug)
putchar('\n');
printf("Sum for row %d = %d\n", r + 1, rowsum);
}
putchar('\n');
for (int c = 0; c < cols; c++)
{
int colsum = 0;
for (int r = 0; r < rows; r++)
{
if (debug)
printf(" a[%d][%d] = %d", r, c, array[r][c]);
colsum += array[r][c];
}
if (debug)
putchar('\n');
printf("Sum for col %d = %d\n", c + 1, colsum);
}
}
static int n; // Rows
static int m; // Columns
// Adapted print_sum_rows_cols with size from n, m
// Using n, m is unnecessary coupling (and can lead to erroneous results)
static void bogus_sum_rows_cols(int array[n][m], int rows, int cols)
{
for (int r = 0; r < rows; r++)
{
int rowsum = 0;
for (int c = 0; c < cols; c++)
{
if (debug)
printf(" a[%d][%d] = %d", r, c, array[r][c]);
rowsum += array[r][c];
}
if (debug)
putchar('\n');
printf("Sum for row %d = %d\n", r, rowsum);
}
putchar('\n');
for (int c = 0; c < cols; c++)
{
int colsum = 0;
for (int r = 0; r < rows; r++)
{
if (debug)
printf(" a[%d][%d] = %d", r, c, array[r][c]);
colsum += array[r][c];
}
if (debug)
putchar('\n');
printf("Sum for col %d = %d\n", c, colsum);
}
}
// Original answer from OP - bogus results (debug printing added).
// Using n, m is unnecessary coupling (and can lead to erroneous results)
// Using i, j as arguments is bogus; the values passed in are ignored
// The 'sum column' loops are completely bogus. The outer loop should
// iterate over columns [0..c) and the inner loop over rows [0..r).
// See print_sum_rows_cols() for the correct (minimal) interface.
static void sum_row_column(int array[n][m], int r, int c, int i, int j)
{
int sumOfColumn = 0;
int sumOfRow = 0;
printf("\n");
for (i = 0; i < r; i++)
{
sumOfRow = 0;
for (j = 0; j < c; j++)
{
if (debug)
printf(" a[%d][%d] = %d", i, j, array[i][j]);
sumOfRow += array[i][j];
}
if (debug)
putchar('\n');
printf("Sum of %d.row=%d\n", i + 1, sumOfRow);
}
printf("\n");
for (j = 0; j < r; j++)
{
sumOfColumn = 0;
for (i = 0; i < c; i++)
{
if (debug)
printf(" a[%d][%d] = %d", i, j, array[i][j]);
sumOfColumn += array[i][j];
}
if (debug)
putchar('\n');
printf("Sum of %d.column=%d\n", j + 1, sumOfColumn);
}
}
// Semi-fixed answer from OP - bogus results (debug printing added).
// Using n, m is unnecessary coupling (and can lead to erroneous results)
// Using i, j as arguments is bogus; the values passed in are ignored
// The 'sum column' loops are completely bogus. The outer loop should
// iterate over columns [0..c) and the inner loop over rows [0..r).
// See print_sum_rows_cols() for the correct (minimal) interface.
static void bogus_sum_row_column(int array[n][m], int r, int c, int i, int j)
{
int sumOfColumn = 0;
int sumOfRow = 0;
printf("\n");
for (i = 0; i < r; i++)
{
sumOfRow = 0;
for (j = 0; j < c; j++)
{
if (debug)
printf(" a[%d][%d] = %d", i, j, array[i][j]);
sumOfRow += array[i][j];
}
if (debug)
putchar('\n');
printf("Sum of %d.row=%d\n", i + 1, sumOfRow);
}
printf("\n");
for (j = 0; j < c; j++) // c, not r
{
sumOfColumn = 0;
for (i = 0; i < r; i++) // r, not c
{
if (debug)
printf(" a[%d][%d] = %d", i, j, array[i][j]);
sumOfColumn += array[i][j];
}
if (debug)
putchar('\n');
printf("Sum of %d.column=%d\n", j + 1, sumOfColumn);
}
}
static void dump_matrix(const char *tag, int rows, int cols, int matrix[rows][cols])
{
printf("Matrix %s (%dx%d):\n", tag, rows, cols);
for (int r = 0; r < rows; r++)
{
for (int c = 0; c < cols; c++)
printf("%3d", matrix[r][c]);
putchar('\n');
}
}
static void test_sequence(int v_debug, int v_n, int v_m, const char *tag,
int rows, int cols, int matrix[rows][cols])
{
debug = v_debug;
n = v_n;
m = v_m;
dump_matrix(tag, rows, cols, matrix);
printf("\nprint_sum_rows_cols():\n");
print_sum_rows_cols(rows, cols, matrix);
printf("\nbogus_sum_rows_cols() - (n = %d, m = %d):\n", n, m);
bogus_sum_rows_cols(matrix, rows, cols);
printf("\nsum_row_column()(n = %d, m = %d):\n", n, m);
sum_row_column(matrix, rows, cols, -1, -1);
printf("\nbogus_sum_row_column()(n = %d, m = %d):\n", n, m);
bogus_sum_row_column(matrix, rows, cols, -1, -1);
putchar('\n');
}
static void test_matrix_summation(const char *tag, int rows, int cols, int matrix[rows][cols])
{
test_sequence(1, 2, 2, tag, rows, cols, matrix);
test_sequence(0, rows, cols, tag, rows, cols, matrix);
}
int main(void)
{
int a1[3][4] =
{
{ 68, 78, 50, 46, },
{ 64, 12, 47, 1, },
{ 86, 10, 84, 62, },
};
int a2[5][3] =
{
{ 4, 30, 19, },
{ 79, 58, 20, },
{ 95, 12, 24, },
{ 20, 37, 72, },
{ 17, 0, 53, },
};
int a3[3][3] =
{
{ 96, 84, 13, },
{ 63, 29, 80, },
{ 97, 98, 48, },
};
test_matrix_summation("A1", 3, 4, a1);
test_matrix_summation("A2", 5, 3, a2);
test_matrix_summation("A3", 3, 3, a3);
return 0;
}
显示的代码输出为：
Matrix A1 (3x4):
68 78 50 46
64 12 47 1
86 10 84 62
print_sum_rows_cols():
a[0][0] = 68 a[0][1] = 78 a[0][2] = 50 a[0][3] = 46
Sum for row 1 = 242
a[1][0] = 64 a[1][1] = 12 a[1][2] = 47 a[1][3] = 1
Sum for row 2 = 124
a[2][0] = 86 a[2][1] = 10 a[2][2] = 84 a[2][3] = 62
Sum for row 3 = 242
a[0][0] = 68 a[1][0] = 64 a[2][0] = 86
Sum for col 1 = 218
a[0][1] = 78 a[1][1] = 12 a[2][1] = 10
Sum for col 2 = 100
a[0][2] = 50 a[1][2] = 47 a[2][2] = 84
Sum for col 3 = 181
a[0][3] = 46 a[1][3] = 1 a[2][3] = 62
Sum for col 4 = 109
bogus_sum_rows_cols() - (n = 2, m = 2):
a[0][0] = 68 a[0][1] = 78 a[0][2] = 50 a[0][3] = 46
Sum for row 0 = 242
a[1][0] = 50 a[1][1] = 46 a[1][2] = 64 a[1][3] = 12
Sum for row 1 = 172
a[2][0] = 64 a[2][1] = 12 a[2][2] = 47 a[2][3] = 1
Sum for row 2 = 124
a[0][0] = 68 a[1][0] = 50 a[2][0] = 64
Sum for col 0 = 182
a[0][1] = 78 a[1][1] = 46 a[2][1] = 12
Sum for col 1 = 136
a[0][2] = 50 a[1][2] = 64 a[2][2] = 47
Sum for col 2 = 161
a[0][3] = 46 a[1][3] = 12 a[2][3] = 1
Sum for col 3 = 59
sum_row_column()(n = 2, m = 2):
a[0][0] = 68 a[0][1] = 78 a[0][2] = 50 a[0][3] = 46
Sum of 1.row=242
a[1][0] = 50 a[1][1] = 46 a[1][2] = 64 a[1][3] = 12
Sum of 2.row=172
a[2][0] = 64 a[2][1] = 12 a[2][2] = 47 a[2][3] = 1
Sum of 3.row=124
a[0][0] = 68 a[1][0] = 50 a[2][0] = 64 a[3][0] = 47
Sum of 1.column=229
a[0][1] = 78 a[1][1] = 46 a[2][1] = 12 a[3][1] = 1
Sum of 2.column=137
a[0][2] = 50 a[1][2] = 64 a[2][2] = 47 a[3][2] = 86
Sum of 3.column=247
bogus_sum_row_column()(n = 2, m = 2):
a[0][0] = 68 a[0][1] = 78 a[0][2] = 50 a[0][3] = 46
Sum of 1.row=242
a[1][0] = 50 a[1][1] = 46 a[1][2] = 64 a[1][3] = 12
Sum of 2.row=172
a[2][0] = 64 a[2][1] = 12 a[2][2] = 47 a[2][3] = 1
Sum of 3.row=124
a[0][0] = 68 a[1][0] = 50 a[2][0] = 64
Sum of 1.column=182
a[0][1] = 78 a[1][1] = 46 a[2][1] = 12
Sum of 2.column=136
a[0][2] = 50 a[1][2] = 64 a[2][2] = 47
Sum of 3.column=161
a[0][3] = 46 a[1][3] = 12 a[2][3] = 1
Sum of 4.column=59
Matrix A1 (3x4):
68 78 50 46
64 12 47 1
86 10 84 62
print_sum_rows_cols():
Sum for row 1 = 242
Sum for row 2 = 124
Sum for row 3 = 242
Sum for col 1 = 218
Sum for col 2 = 100
Sum for col 3 = 181
Sum for col 4 = 109
bogus_sum_rows_cols() - (n = 3, m = 4):
Sum for row 0 = 242
Sum for row 1 = 124
Sum for row 2 = 242
Sum for col 0 = 218
Sum for col 1 = 100
Sum for col 2 = 181
Sum for col 3 = 109
sum_row_column()(n = 3, m = 4):
Sum of 1.row=242
Sum of 2.row=124
Sum of 3.row=242
Sum of 1.column=222
Sum of 2.column=130
Sum of 3.column=200
bogus_sum_row_column()(n = 3, m = 4):
Sum of 1.row=242
Sum of 2.row=124
Sum of 3.row=242
Sum of 1.column=218
Sum of 2.column=100
Sum of 3.column=181
Sum of 4.column=109
Matrix A2 (5x3):
4 30 19
79 58 20
95 12 24
20 37 72
17 0 53
print_sum_rows_cols():
a[0][0] = 4 a[0][1] = 30 a[0][2] = 19
Sum for row 1 = 53
a[1][0] = 79 a[1][1] = 58 a[1][2] = 20
Sum for row 2 = 157
a[2][0] = 95 a[2][1] = 12 a[2][2] = 24
Sum for row 3 = 131
a[3][0] = 20 a[3][1] = 37 a[3][2] = 72
Sum for row 4 = 129
a[4][0] = 17 a[4][1] = 0 a[4][2] = 53
Sum for row 5 = 70
a[0][0] = 4 a[1][0] = 79 a[2][0] = 95 a[3][0] = 20 a[4][0] = 17
Sum for col 1 = 215
a[0][1] = 30 a[1][1] = 58 a[2][1] = 12 a[3][1] = 37 a[4][1] = 0
Sum for col 2 = 137
a[0][2] = 19 a[1][2] = 20 a[2][2] = 24 a[3][2] = 72 a[4][2] = 53
Sum for col 3 = 188
bogus_sum_rows_cols() - (n = 2, m = 2):
a[0][0] = 4 a[0][1] = 30 a[0][2] = 19
Sum for row 0 = 53
a[1][0] = 19 a[1][1] = 79 a[1][2] = 58
Sum for row 1 = 156
a[2][0] = 58 a[2][1] = 20 a[2][2] = 95
Sum for row 2 = 173
a[3][0] = 95 a[3][1] = 12 a[3][2] = 24
Sum for row 3 = 131
a[4][0] = 24 a[4][1] = 20 a[4][2] = 37
Sum for row 4 = 81
a[0][0] = 4 a[1][0] = 19 a[2][0] = 58 a[3][0] = 95 a[4][0] = 24
Sum for col 0 = 200
a[0][1] = 30 a[1][1] = 79 a[2][1] = 20 a[3][1] = 12 a[4][1] = 20
Sum for col 1 = 161
a[0][2] = 19 a[1][2] = 58 a[2][2] = 95 a[3][2] = 24 a[4][2] = 37
Sum for col 2 = 233
sum_row_column()(n = 2, m = 2):
a[0][0] = 4 a[0][1] = 30 a[0][2] = 19
Sum of 1.row=53
a[1][0] = 19 a[1][1] = 79 a[1][2] = 58
Sum of 2.row=156
a[2][0] = 58 a[2][1] = 20 a[2][2] = 95
Sum of 3.row=173
a[3][0] = 95 a[3][1] = 12 a[3][2] = 24
Sum of 4.row=131
a[4][0] = 24 a[4][1] = 20 a[4][2] = 37
Sum of 5.row=81
a[0][0] = 4 a[1][0] = 19 a[2][0] = 58
Sum of 1.column=81
a[0][1] = 30 a[1][1] = 79 a[2][1] = 20
Sum of 2.column=129
a[0][2] = 19 a[1][2] = 58 a[2][2] = 95
Sum of 3.column=172
a[0][3] = 79 a[1][3] = 20 a[2][3] = 12
Sum of 4.column=111
a[0][4] = 58 a[1][4] = 95 a[2][4] = 24
Sum of 5.column=177
bogus_sum_row_column()(n = 2, m = 2):
a[0][0] = 4 a[0][1] = 30 a[0][2] = 19
Sum of 1.row=53
a[1][0] = 19 a[1][1] = 79 a[1][2] = 58
Sum of 2.row=156
a[2][0] = 58 a[2][1] = 20 a[2][2] = 95
Sum of 3.row=173
a[3][0] = 95 a[3][1] = 12 a[3][2] = 24
Sum of 4.row=131
a[4][0] = 24 a[4][1] = 20 a[4][2] = 37
Sum of 5.row=81
a[0][0] = 4 a[1][0] = 19 a[2][0] = 58 a[3][0] = 95 a[4][0] = 24
Sum of 1.column=200
a[0][1] = 30 a[1][1] = 79 a[2][1] = 20 a[3][1] = 12 a[4][1] = 20
Sum of 2.column=161
a[0][2] = 19 a[1][2] = 58 a[2][2] = 95 a[3][2] = 24 a[4][2] = 37
Sum of 3.column=233
Matrix A2 (5x3):
4 30 19
79 58 20
95 12 24
20 37 72
17 0 53
print_sum_rows_cols():
Sum for row 1 = 53
Sum for row 2 = 157
Sum for row 3 = 131
Sum for row 4 = 129
Sum for row 5 = 70
Sum for col 1 = 215
Sum for col 2 = 137
Sum for col 3 = 188
bogus_sum_rows_cols() - (n = 5, m = 3):
Sum for row 0 = 53
Sum for row 1 = 157
Sum for row 2 = 131
Sum for row 3 = 129
Sum for row 4 = 70
Sum for col 0 = 215
Sum for col 1 = 137
Sum for col 2 = 188
sum_row_column()(n = 5, m = 3):
Sum of 1.row=53
Sum of 2.row=157
Sum of 3.row=131
Sum of 4.row=129
Sum of 5.row=70
Sum of 1.column=178
Sum of 2.column=100
Sum of 3.column=63
Sum of 4.column=194
Sum of 5.column=107
bogus_sum_row_column()(n = 5, m = 3):
Sum of 1.row=53
Sum of 2.row=157
Sum of 3.row=131
Sum of 4.row=129
Sum of 5.row=70
Sum of 1.column=215
Sum of 2.column=137
Sum of 3.column=188
Matrix A3 (3x3):
96 84 13
63 29 80
97 98 48
print_sum_rows_cols():
a[0][0] = 96 a[0][1] = 84 a[0][2] = 13
Sum for row 1 = 193
a[1][0] = 63 a[1][1] = 29 a[1][2] = 80
Sum for row 2 = 172
a[2][0] = 97 a[2][1] = 98 a[2][2] = 48
Sum for row 3 = 243
a[0][0] = 96 a[1][0] = 63 a[2][0] = 97
Sum for col 1 = 256
a[0][1] = 84 a[1][1] = 29 a[2][1] = 98
Sum for col 2 = 211
a[0][2] = 13 a[1][2] = 80 a[2][2] = 48
Sum for col 3 = 141
bogus_sum_rows_cols() - (n = 2, m = 2):
a[0][0] = 96 a[0][1] = 84 a[0][2] = 13
Sum for row 0 = 193
a[1][0] = 13 a[1][1] = 63 a[1][2] = 29
Sum for row 1 = 105
a[2][0] = 29 a[2][1] = 80 a[2][2] = 97
Sum for row 2 = 206
a[0][0] = 96 a[1][0] = 13 a[2][0] = 29
Sum for col 0 = 138
a[0][1] = 84 a[1][1] = 63 a[2][1] = 80
Sum for col 1 = 227
a[0][2] = 13 a[1][2] = 29 a[2][2] = 97
Sum for col 2 = 139
sum_row_column()(n = 2, m = 2):
a[0][0] = 96 a[0][1] = 84 a[0][2] = 13
Sum of 1.row=193
a[1][0] = 13 a[1][1] = 63 a[1][2] = 29
Sum of 2.row=105
a[2][0] = 29 a[2][1] = 80 a[2][2] = 97
Sum of 3.row=206
a[0][0] = 96 a[1][0] = 13 a[2][0] = 29
Sum of 1.column=138
a[0][1] = 84 a[1][1] = 63 a[2][1] = 80
Sum of 2.column=227
a[0][2] = 13 a[1][2] = 29 a[2][2] = 97
Sum of 3.column=139
bogus_sum_row_column()(n = 2, m = 2):
a[0][0] = 96 a[0][1] = 84 a[0][2] = 13
Sum of 1.row=193
a[1][0] = 13 a[1][1] = 63 a[1][2] = 29
Sum of 2.row=105
a[2][0] = 29 a[2][1] = 80 a[2][2] = 97
Sum of 3.row=206
a[0][0] = 96 a[1][0] = 13 a[2][0] = 29
Sum of 1.column=138
a[0][1] = 84 a[1][1] = 63 a[2][1] = 80
Sum of 2.column=227
a[0][2] = 13 a[1][2] = 29 a[2][2] = 97
Sum of 3.column=139
Matrix A3 (3x3):
96 84 13
63 29 80
97 98 48
print_sum_rows_cols():
Sum for row 1 = 193
Sum for row 2 = 172
Sum for row 3 = 243
Sum for col 1 = 256
Sum for col 2 = 211
Sum for col 3 = 141
bogus_sum_rows_cols() - (n = 3, m = 3):
Sum for row 0 = 193
Sum for row 1 = 172
Sum for row 2 = 243
Sum for col 0 = 256
Sum for col 1 = 211
Sum for col 2 = 141
sum_row_column()(n = 3, m = 3):
Sum of 1.row=193
Sum of 2.row=172
Sum of 3.row=243
Sum of 1.column=256
Sum of 2.column=211
Sum of 3.column=141
bogus_sum_row_column()(n = 3, m = 3):
Sum of 1.row=193
Sum of 2.row=172
Sum of 3.row=243
Sum of 1.column=256
Sum of 2.column=211
Sum of 3.column=141
您可以在打开或关闭调试时调整以获得不同的输出量，但仔细审查结果会显示main()程序中定义的矩阵的实际行和列总和的差异
当全局变量与参数同步时，所提出的答案在方形矩阵上获得“正确”答案的事实说明了为什么进行全面测试很重要。
如果代码只是为了处理方形矩阵，那么您不会同时拥有n和m（并且您不需要r和c ）。但是你不应该把函数绑定到全局变量都是;您应该定义r和c（如果矩阵必须是正方形，则只定义其中一个）。
这也说明了为什么MCVE（Minimal, Complete, Verifiable Example）很重要。变量n和m中的值未记录在问题或建议的答案中，因此有必要梳理出它们所代表的含义。
答案 1 :(得分：-1)
我做了主要的;
case 8:
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
printf("Enter %d.row and %d.column: ",i+1,j+1);
scanf("%d",&array[i][j]);
}
}
sum_row_column(array,n,m,i,j);
break;
和功能;
void sum_row_column(int array[n][m],int r,int c,int i,int j)
{
int sumOfColumn=0;
int sumOfRow=0;
printf("\n");
for(i=0;i<r;i++)
{
sumOfRow=0;
for(j=0;j<c;j++)
{
sumOfRow+=array[i][j];
}
printf("Sum of %d.row=%d\n",i+1,sumOfRow);
}
printf("\n");
for(j=0;j<r;j++)
{
sumOfColumn=0;
for(i=0;i<c;i++)
{
sumOfColumn+=array[i][j];
}
printf("Sum of %d.column=%d\n",j+1,sumOfColumn);
}
}
及其工作！谢谢你的帮助。