如何从数组中的输入数据文件中打印值？
如何从数组中的输入数据文件中打印值？
提问于 2014-10-18 15:29:58
我是一个完全的初学者程序员，所以请容忍我。
因此，我有一个输入文本文件，我将使用它作为程序的输入，在命令窗口中使用program.exe < data.txt。文本文件有5行，每一行有3个双值，比如30.0、70.0、0.05等等。
我想基本上使用一个结构数组来打印这些输入值，比如printf(“第一个值是%f"，arrayi)。
以下是我目前不正确的代码：
#include <stdio.h>
#include <stdlib.h>
#define MAXSOURCES 100
typedef struct {
double x;
double y;
} coordinates_t;
typedef struct {
coordinates_t point;
double w;
} soundsource_t;
coordinates_t a;
soundsource_t b;
int main(int argc, char *argv[]) {
int i;
while(scanf("%lf %lf %lf", &a.x, &a.y, &b.w) == 3) {
soundsource_t soundsource[MAXSOURCES][2];
for (i = 0; i <= MAXSOURCES; i++) {
printf("%d", soundsource[i][0]);
printf("%d", soundsource[i][1]);
printf("%d", soundsource[i][2]);
printf("\n");
}
}
return 0;
}有人能帮我修好密码吗？谢谢
回答 2
Stack Overflow用户
回答已采纳
发布于 2014-10-18 16:00:00
样品修复
int main(int argc, char *argv[]) {
soundsource_t soundsource[MAXSOURCES];
int i, n = 0;
while(scanf("%lf %lf %lf", &a.x, &a.y, &b.w) == 3) {
soundsource[n].point = a;
soundsource[n].w = b.w;
n += 1;
}
for (i = 0; i < n; i++) {
printf("%f,", soundsource[i].point.x);
printf("%f ", soundsource[i].point.y);
printf("%f\n",soundsource[i].w);
}
return 0;
}Stack Overflow用户
发布于 2014-10-18 16:12:13
确保你读到了第一个评论。但是，您似乎还需要更多的研究，所以我将给您一个如何修改代码以实现这一目标的示例。
#include <stdio.h>
#include <stdlib.h>
#define MAXSOURCES 100
typedef struct {
double x;
double y;
} coordinates_t;
typedef struct {
coordinates_t point;
double w;
} soundsource_t;
int main(int argc, char *argv[]) {
int i, n = 0; // i is a counter and n will be the number of sources we actually read
// we declare an array of sources, with size MAXSOURCES
soundsource_t array[MAXSOURCES];
/*
* Now every cell of the array is a soundsource_t struct.
* So, every cell is capable of storing a 'point' and a 'w'.
*/
// We will read from stdin, three doubles per loop
// and we will store them in the n-th cell of the array
while(scanf("%lf %lf %lf", &(array[n].point.x), &(array[n].point.y), &(array[n].w)) == 3) {
n++; // increment n, so that we keep track of number of sources we read
if(n == MAXSOURCES) // we reached the limit, so stop reading
break;
}
// print the sources we read. Notice that we go until n and not MAXSOURCES,
// since we read n sources.
for(i = 0; i < n; ++i) {
// print the i-th cell
printf("%lf %lf %lf\n", array[i].point.x, array[i].point.y, array[i].w);
}
return 0;
}输出：
0.1 0.2 0.3
0.7 0.8 0.9
e <-- here I typed a letter to stop reading from input
0.100000 0.200000 0.300000
0.700000 0.800000 0.900000页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/26441387
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