从txt文件创建int数组,在C中使用多行和数字之间的空格
我有一个数独棋盘。我需要将sudoku.txt的数字捕获到C中的整数数组中 我的sudoku.txt是多行的,数字之间用空格分隔从txt文件创建int数组,在C中使用多行和数字之间的空格,c,arrays,file,sudoku,C,Arrays,File,Sudoku,我有一个数独棋盘。我需要将sudoku.txt的数字捕获到C中的整数数组中 我的sudoku.txt是多行的,数字之间用空格分隔 6 2 4 5 3 9 1 8 7 5 1 9 7 2 8 6 3 4 8 3 7 6 1 4 2 9 5 1 4 3 8 6 5 7 2 9 9 5 8 2 4 7 3 6 1 7 6 2 3 9 1 4 5 8 3 7 1 9 5 6 8 4 2 4 9 6 1 8 2 5 7 3 2 8 5 4 7 3 9 1 6 我想像这样加载数组中的所有数字 例如: 我见
6 2 4 5 3 9 1 8 7
5 1 9 7 2 8 6 3 4
8 3 7 6 1 4 2 9 5
1 4 3 8 6 5 7 2 9
9 5 8 2 4 7 3 6 1
7 6 2 3 9 1 4 5 8
3 7 1 9 5 6 8 4 2
4 9 6 1 8 2 5 7 3
2 8 5 4 7 3 9 1 6
我想像这样加载数组中的所有数字
例如:
我见过很多txt到INT数组的帖子,比如12345678或12345678一行。
但多行和分隔还没有
代码:
即使您确实检查了文件是否已打开,但在失败时您仍然继续读取,这是错误的,此代码不会以这种方式失败,但我认为它也不会工作,因为您的文件显然不存在 另外,在打印循环中,您试图打印比数组中更多的元素,我也解决了这个问题
#include <stdio.h>
#include <stdlib.h>
void sudoku_checker(int N ,int a[])
{
int i,j;
int count;
/* This loop calculate the sum of each row */
/* and check foreach if it equals to (1+2+3+4+5+6+7+8+9) = 45 o not*/
count = 0;
for (i = 0; i < N * N ; i = i + 9)
{
int sumRow = 0;
for (j = i; j < i + 9; j++)
sumRow = sumRow + a[j];
if (sumRow != 45)
count++;
}
/* This loop calculate the sum of each col*/
/* and check foreach if it equals to (1+2+3+4+5+6+7+8+9) = 45 o not*/
for (i = 0; i < N; i++)
{
int sumCol = 0;
for (j = i ; j < N * N; j = j + 9)
sumCol = sumCol + a[j];
if (sumCol != 45)
count++;
}
printf("Fails: %d\n", count);
if (count == 0)
printf("OK\n");
else
printf("Successes\n");
}
int main()
{
//info.txt it's the sudoku file
int i, j, k, arrayofNumbers[81];
FILE *myFile;
int N = 9;
int readCount = 0;
myFile = fopen("file.dat", "r");
if (myFile == NULL)
{
printf("the file could not be opened for reading\n");
/* you should abort the program, if the file was not found */
return -1;
}
while (fscanf(myFile, "%1d", &arrayofNumbers[i]) == 1)
readCount++;
for (j = 0 ; j < readCount ; ++j)
{
printf(" %1d ", arrayofNumbers[j]);
if ((1 + j) % 4 == 0)
printf("\n");
}
sudoku_checker(N,arrayofNumbers);
return 0;
}
请注意,您没有检查fopen成功与否的任何具体原因?另外,请详细说明这不起作用。@iharob检查它是确定的,发布一个试图读取文件然后打印它的完整程序。为什么你让人们很难帮助你?整个计划都没有公布。你的主要职能是什么?你的include指令在哪里?您需要发布一个完整的可编译程序,尽可能简短地演示该问题。没有这个,,你在浪费每个人的时间。如果我使用你的代码,它会说我的文件存在并按如下方式打印数组:0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1606418432 32767-2102815162-332136427-148998249 14956788624-1188721324 167778272 1606416480 32767 16 0 1 0 104 1 1606578576 32767 1606416240 32767 1606423158 32767 1606416264 32767 1检查文件内容,尝试更新的代码,我刚刚注意到,您没有检查是否从文件中读取了值,例如,在leidos,los valores del arreglo quedan sinicializar y vas a obtener lo que llaman valores basura como los que tu program a estáimimimiendo中。它是固定的,并且是有效的。格雷西亚斯·波拉·阿尤达@iharob
//
// main.c
// sudoku
//
// Created by Ramón Serrano López on 23/1/15.
// Copyright (c) 2015 Ramón Serrano López. All rights reserved.
//
#include <stdio.h>
#include <stdlib.h>
void sudoku_checker(int N ,int a[]){
int i,j;
int count=0;
/* This loop calculate the sum of each row */
/* and check foreach if it equals to (1+2+3+4+5+6+7+8+9) = 45 o not*/
for (i = 0; i < N*N; i = i+9) {
int sumRow = 0;
for (j = i; j < i+9; j++) {
sumRow = sumRow + a[j];
}
if (sumRow != 45) {
count++;
}
}
/* This loop calculate the sum of each col*/
/* and check foreach if it equals to (1+2+3+4+5+6+7+8+9) = 45 o not*/
for (i = 0; i < N; i++) {
int sumCol = 0;
for (j = i; j < N*N; j = j+9) {
sumCol = sumCol + a[j];
}
if (sumCol != 45) {
count++;
}
}
printf("Fails: %d\n", count);
if(count==0)
printf("OK\n");
else
printf("Successes\n");
}
int main()
{
//info.txt it's the sudoku file
FILE *myFile = fopen("/Users/ramonserranolopez/Desktop/SO/sudoku/sudoku/info.txt","r");
int i, j, k, arrayofNumbers[81];
if (myFile == NULL) {
printf("the file could not be opened for reading\n");
} else {
for (i=0; i<81; i++) {
fscanf(myFile, "%1d", &arrayofNumbers[i]);
}
}
//provisional array to check if the function sudoku_cheker works
//int a[]={6, 2, 4, 5, 3, 9, 1, 8, 7, 5, 1, 9, 7, 2, 8, 6, 3, 4, 8, 3, 7, 6, 1, 4, 2, 9, 5, 1, 4, 3, 8, 6, 5, 7, 2, 9, 9, 5, 8, 2, 4, 7, 3, 6, 1, 7, 6, 2, 3, 9, 1, 4, 5, 8, 3, 7, 1, 9, 5, 6, 8, 4, 2, 4, 9, 6, 1, 8, 2, 5, 7, 3, 2, 8, 5, 4, 7, 3, 9, 1, 6};
//prints the array
int N = 9;
for (j = 0; j < N*N; j = j+4) {
for (k = j; k < j+4; k++) {
printf(" %1d ", arrayofNumbers[k]);
}
printf("\n");
}
sudoku_checker(N,arrayofNumbers);
return 0;
}
#include <stdio.h>
#include <stdlib.h>
void sudoku_checker(int N ,int a[])
{
int i,j;
int count;
/* This loop calculate the sum of each row */
/* and check foreach if it equals to (1+2+3+4+5+6+7+8+9) = 45 o not*/
count = 0;
for (i = 0; i < N * N ; i = i + 9)
{
int sumRow = 0;
for (j = i; j < i + 9; j++)
sumRow = sumRow + a[j];
if (sumRow != 45)
count++;
}
/* This loop calculate the sum of each col*/
/* and check foreach if it equals to (1+2+3+4+5+6+7+8+9) = 45 o not*/
for (i = 0; i < N; i++)
{
int sumCol = 0;
for (j = i ; j < N * N; j = j + 9)
sumCol = sumCol + a[j];
if (sumCol != 45)
count++;
}
printf("Fails: %d\n", count);
if (count == 0)
printf("OK\n");
else
printf("Successes\n");
}
int main()
{
//info.txt it's the sudoku file
int i, j, k, arrayofNumbers[81];
FILE *myFile;
int N = 9;
int readCount = 0;
myFile = fopen("file.dat", "r");
if (myFile == NULL)
{
printf("the file could not be opened for reading\n");
/* you should abort the program, if the file was not found */
return -1;
}
while (fscanf(myFile, "%1d", &arrayofNumbers[i]) == 1)
readCount++;
for (j = 0 ; j < readCount ; ++j)
{
printf(" %1d ", arrayofNumbers[j]);
if ((1 + j) % 4 == 0)
printf("\n");
}
sudoku_checker(N,arrayofNumbers);
return 0;
}