Does not take or print the declared array C

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Does not take or print the declared array C

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#1 by (1 votes)

1

my problem is the following, I do not know why I do not take the matrix as I declare 3x3 take me and print as if it were a common arrangement, I hope you can help me. Thank you very much!

int main() {

    int mat [3][3];
    int i, x;

    printf("Ingresa el contenido de la matriz: \n");


    for(i=0; i<3; i++) {
        for(x=0; i<3;i++) {

            scanf("%d", &mat[i][x]);

        }
    }

    for(int i=0; i<3; i++) {
        for(int x=0; i<3;i++) {
            printf("%d \t",mat[i][x]);

        }
        printf("\n");
    }

    return 0;
}

c matrices arreglos

asked by Josias99 28.06.2017 в 18:55

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1 answer

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score 1 · Accepted Answer

First of all, your mistake is that in the second for, you compare the first variable and increase it, causing an erroneous behavior.

It should be mentioned that in a for you can not declare and set a variable, ANSI C does not allow it, you must first define it and initialize it in the for.

#include <stdio.h>
int main() 
{
    int i,j, matriz[3][3]; // ANSI C impide declarar variables dentro de un for

    puts("Ingresa el contenido de la matriz");    

    for (i = 0; i < 3; i++) {
        for (j = 0; j < 3; j++) { // nota que se debe incrementar j y no i
            scanf("%d", &matriz[i][j]);
        }
    }

    for(i = 0; i < 3; i++) {
        for(j = 0; j < 3; j++) { // lo mismo, se incrementa j y no i
            printf("%d \t",matriz[i][j]);
        }
        puts("");
    }

    return 0;
}