use*_*383 3 c declaration matrix
我正在自学(几天前开始)并且无法用葡萄牙语(我的母语)找到可靠的信息.在太头疼之后,我放弃了来到这里寻求帮助.我不知道如何制作阵列/二维阵列的双打,只有汉堡.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;}
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输出与预期一致:
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
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**
**
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;}
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输出总是错误的,因为我一直在改变它.我宣布长双打,双打,整数和双打等.每次代码输出错误的答案.某些组合输出负数.其他人只是输出ZERO到处都是.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
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要么
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
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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?
%dprintf期望int参数的格式说明符,但是你传递了一个double.然后您的程序会显示未定义的行为.有关如何输出类型值的信息,请参阅您最喜欢的C或C++教科书double.