C语言 在 C 中创建一个基本矩阵（由用户输入！）
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Create a basic matrix in C (input by user !)
提问by NLed
I'm trying to ask the user to enter the number of columns and rows they want in a matrix, and then enter the values in the matrix... I'm going to let them insert numbers one row at a time.
我试图让用户在矩阵中输入他们想要的列数和行数，然后在矩阵中输入值......我将让他们一次插入一行数字。
How can I create such function ?
我怎样才能创建这样的功能？
#include<stdio.h>
main(){
int mat[10][10],i,j;
for(i=0;i<2;i++)
for(j=0;j<2;j++){
scanf("%d",&mat[i][j]);
}
for(i=0;i<2;i++)
for(j=0;j<2;j++)
printf("%d",mat[i][j]);
}
This works for entering the numbers, but it displays them all in one line... The issue here is that I don't know how many columns or rows the user wants, so I cant print out %d %d %d in a matrix form...
这适用于输入数字，但它将它们全部显示在一行中......这里的问题是我不知道用户想要多少列或行，所以我无法打印出 %d %d %d矩阵形式...
Any thoughts?
有什么想法吗？
Thanks :)
谢谢 ：）
回答by Jacob
How about the following?
以下情况如何？
First ask the user for the number of rows and columns, store that in say, nrowsand ncols(i.e. scanf("%d", &nrows);) and then allocate memory for a 2D arrayof size nrows x ncols. Thus you can have a matrix of a size specified by the user, and not fixed at some dimension you've hardcoded!
首先询问用户行数和列数，将其存储在 say, nrowsand ncols(ie scanf("%d", &nrows);) 中，然后为大小为nrows x ncols的二维数组分配内存。因此，您可以拥有一个由用户指定大小的矩阵，而不是固定在您硬编码的某个维度上！
Then store the elements with for(i = 0;i < nrows; ++i) ...and display the elements in the same way except you throw in newlines after every row, i.e.
然后for(i = 0;i < nrows; ++i) ...以相同的方式存储元素并显示元素，除了在每一行后添加换行符，即
for(i = 0; i < nrows; ++i)
{
for(j = 0; j < ncols ; ++j)
{
printf("%d\t",mat[i][j]);
}
printf("\n");
}
回答by Yann Ramin
You need to dynamically allocate your matrix. For instance:
您需要动态分配矩阵。例如：
int* mat;
int dimx,dimy;
scanf("%d", &dimx);
scanf("%d", &dimy);
mat = malloc(dimx * dimy * sizeof(int));
This creates a linear array which can hold the matrix. At this point you can decide whether you want to access it column or row first. I would suggest making a quick macro which calculates the correct offset in the matrix.
这将创建一个可以容纳矩阵的线性阵列。此时，您可以决定是要先访问它的列还是行。我建议制作一个快速宏来计算矩阵中的正确偏移量。
回答by Keith Nicholas
need a
需要一个
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
printf("%d",mat[i][j]);
}
printf("\n");
}
回答by jeff
#include<stdio.h>
int main(void)
{
int mat[10][10],i,j;
printf("Enter your matrix\n");
for(i=0;i<2;i++)
for(j=0;j<2;j++)
{
scanf("%d",&mat[i][j]);
}
printf("\nHere is your matrix:\n");
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
printf("%d ",mat[i][j]);
}
printf("\n");
}
}
回答by TonuSree
This is my answer
这是我的回答
#include<stdio.h>
int main()
{int mat[100][100];
int row,column,i,j;
printf("enter how many row and colmn you want:\n \n");
scanf("%d",&row);
scanf("%d",&column);
printf("enter the matrix:");
for(i=0;i<row;i++){
for(j=0;j<column;j++){
scanf("%d",&mat[i][j]);
}
printf("\n");
}
for(i=0;i<row;i++){
for(j=0;j<column;j++){
printf("%d \t",mat[i][j]);}
printf("\n");}
}
I just choose an approximate value for the row and column. My selected row or column will not cross the value.and then I scan the matrix element then make it in matrix size.
我只是为行和列选择一个近似值。我选择的行或列不会与值交叉。然后我扫描矩阵元素，然后将其设为矩阵大小。
回答by Leonardo Amadori Lima
int rows, cols , i, j;
printf("Enter number of rows and cols for the matrix: \n");
scanf("%d %d",&rows, &cols);
int mat[rows][cols];
printf("enter the matrix:");
for(i = 0; i < rows ; i++)
for(j = 0; j < cols; j++)
scanf("%d", &mat[i][j]);
printf("\nThe Matrix is:\n");
for(i = 0; i < rows ; i++)
{
for(j = 0; j < cols; j++)
{
printf("%d",mat[i][j]);
printf("\t");
}
printf("\n");
}
}
}
回答by CHECK HAROUN
//R stands for ROW and C stands for COLUMN:
//i stands for ROW and j stands for COLUMN:
#include<stdio.h>
int main(){
int M[100][100];
int R,C,i,j;
printf("Please enter how many rows you want:\n");
scanf("%d",& R);
printf("Please enter how column you want:\n");
scanf("%d",& C);
printf("Please enter your matrix:\n");
for(i = 0; i < R; i++){
for(j = 0; j < C; j++){
scanf("%d", &M[i][j]);
}
printf("\n");
}
for(i = 0; i < R; i++){
for(j = 0; j < C; j++){
printf("%d\t", M[i][j]);
}
printf("\n");
}
getch();
return 0;
}
