如何在 C 中的矩阵(二维数组)内声明双精度数?

How to declare doubles inside a matrix (2d arrays) in C?

我正在自学(几天前开始),找不到可靠的葡萄牙语(我的母语)信息。 头痛太多之后,我放弃了,来到这里寻求帮助。 我不知道如何制作 array/2d 双精度数组,只有整数。 以下代码按预期工作:

int main(void){    
int matriz[5][2] =  {{1,2},  // DECLARATION OF MATRIX COMPOSED OF INTERGERS
                     {3,4},
                     {5,6},
                     {7,8},
                     {9,10}};
int maior = matriz[0][0];   // BIGGEST NUMBER INSIDE MATRIX, STARTED AS FIRST POSITION
int soma = 0;               // SUM OF NUMBERS START WITH ZERO
int lin = 0;                // INDICATION OF LINE
int col = 0;                // INDICATION OF COLUMN
for (lin = 0; lin < 5; lin++){
    for (col = 0; col < 2; col++){
    if (maior < matriz[lin][col]) {maior = matriz[lin][col];}
    soma += matriz[lin][col];
    }   
}
printf("A soma dos numeros eh %d e o maior numero eh %d\n", soma, maior); 
// THE SUM OF NUMBERS IS %D AND THE BIGGEST NUMBER IS %D
// I COPIED THE FOLLOWING CODE FROM ELSEWHERE TO SEE WHAT WAS GOING ON
    int i, j;
    for ( i = 0; i < 5; i++ ) {
          for ( j = 0; j < 2; j++ ) {
         printf("a[%d][%d] = %d\n", i,j, matriz[i][j] );
      }
   }
return 0;} 

输出符合预期:

A soma dos numeros eh 55 e o maior numero eh 10
a[0][0] = 1 
a[0][1] = 2 
a[1][0] = 3 
a[1][1] = 4 
a[2][0] = 5 
a[2][1] = 6 
a[3][0] = 7 
a[3][1] = 8
a[4][0] = 9
a[4][1] = 10

**

然而,任何更改如以下代码总是错误的:

**

int main(void){
double matriz2[5][2] = {{10000,1000},
                        {100,10},
                        {1,0.1},
                        {0.01,0.001},
                        {0.0001,0.00001}};
double maior2 = matriz2[0][0];
double soma2 = 0;
int lin2 = 0;
int col2 = 0;
for (lin2 = 0; lin2 < 5; lin2++){
    for (col2 = 0; col2 < 2; col2++){
    if (maior2 < matriz2[lin2][col2]) {maior2 = matriz2[lin2][col2];}
    soma2 += matriz2[lin2][col2];
    }   
}
printf("A soma dos numeros eh %d e o maior numero eh %d\n", soma2, maior2);

for ( i = 0; i < 5; i++ ) {
      for ( j = 0; j < 2; j++ ) {
         printf("a[%d][%d] = %d\n", i,j, matriz2[i][j] );
      }
   }
return 0;} 

输出总是错误的,因为我一直在改变它。 我声明了长双打、双打、整数和双打等。每次代码都输出错误的答案。 一些组合输出负数。其他人只是到处输出零。两个例子:

A soma dos numeros eh 0 e o maior numero eh 0
a[0][0] = 10000
a[0][1] = 1000
a[1][0] = 100
a[1][1] = 10
a[2][0] = 1
a[2][1] = 0
a[3][0] = 0
a[3][1] = 0
a[4][0] = 0
a[4][1] = 0

A soma dos numeros eh 953826337 e o maior numero eh 0
a[0][0] = 0
a[0][1] = 0
a[1][0] = 0
a[1][1] = 0
a[2][0] = 0
a[2][1] = -1717986918
a[3][0] = 1202590843
a[3][1] = -755914244
a[4][0] = -350469331
a[4][1] = -1998362383

我根本找不到在矩阵中声明或放置双数的方法。 我该怎么做?

%d printf 的格式说明符需要一个 int 参数,但您传递的是 double。然后您的程序会表现出未定义的行为。查阅您最喜欢的 C 或 C++ 教科书,了解如何输出 double.

类型的值

您已经标记了 CC++,您应该选择一个。我会给出两种语言的答案。


如果将 Cprintf() 一起使用,当您将二维数组从 int 更改为 double 时,计算似乎没问题,但您的输出有问题printf()。您正在使用 %d,它需要一个 int,但您将其传递给一个 double,这会导致未定义的行为。使用 floatdouble 时,将其更改为 %f%g


如果使用 C++ 尽量不要使用 printf()。相反,您应该包括 iostream。然后你的输出看起来像:

// printf("A soma dos numeros eh %d e o maior numero eh %d\n", soma2, maior2);
std::cout << "A soma dos numerous eh " << soma2 << " e o maior numero eh << maior2 << '\n';

for ( i = 0; i < 5; i++ ) {
    for ( j = 0; j < 2; j++ ) {
        // printf("a[%d][%d] = %d\n", i,j, matriz2[i][j] );
        std::cout << "a[" << i << "][" << j "] = " << matriz2[i][j] << '\n';
    }
}