我正在尝试用命令传递的3个档案中的值读取构造。我有三种类型的硬币，它们的价格和日期已变成数组，但不是动态的。我该如何从结构上调整这些日期和价格？
命令
a.exe BTC.csv NEO.csv IOT.csv
btc.csv
253
02/20/18,11403.7
02/19/18,11225.3
02/18/18,10551.8
02/17/18,11112.7
02/16/18,10233.9
...
neo.csv
253
02/20/18,128.36
02/19/18,137.47
02/18/18,127.38
02/17/18,136.75
02/16/18,128.85
...
iot.csv
253
2/20/18,1.91
2/19/18,2.09
2/18/18,1.98
2/17/18,2.2
2/16/18,2.1
...
代码：
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
typedef struct
{
int month;
int day;
int year;
double value;
}array;
typedef struct
{ //I want allocate these arrays dynamically
int month[253];
int day[253];
int year[253];
double newvalue[253];
}coin;
int main(int argc, char *argv[]){
FILE *csv;
char string[9];
long int n_lines=0;
int n = argc-1;
coin *m = (coin*)malloc(n*sizeof(coin));
for(int z=1; z<argc; z++)
{
sprintf(string, "%s", argv[z]);
if((csv=fopen(string, "r")) == NULL)
{
printf("%s not foundn", string);
exit(1);
}
fscanf(csv, "%li", &n_lines); //n_lines=253
array *v;
v=(array*)malloc(n_lines*sizeof(array));
char line[256];
int i=0;
while (fgets(line, 256, csv) != NULL && i<n_lines)
{
int count = fscanf(csv, "%d/%d/%d,%lf", &v[i].month, &v[i].day, &v[i].year, &v[i].value);
m[z-1].month[i] = v[i].month;
m[z-1].day[i] = v[i].day;
m[z-1].year[i] = v[i].year;
m[z-1].newvalue[i] = (v[i].value)/2;
i++;
}
free(v);
fclose(csv);
}
for(int z=1;i<argc;z++)
{
for(int i=0;i<n_lines;i++)
{
printf("%0.2d/%0.2d/%0.2d %lfn", m[z-1].month[i], m[z-1].day[i], m[z-1].year[i], m[z-1].newvalue[i]);
}
}
return 0;
}
，而不是将struct设置为一堆数组：
typedef struct
{ //I want allocate these arrays dynamically
int month[253];
int day[253];
int year[253];
double newvalue[253];
}coin;
只使用指针：
typedef struct
{
int *month;
int *day;
int *year;
double *newvalue;
} coin;
并在v=(array*)malloc(n_lines*sizeof(array));之后动态分配内存：
coin c;
c.month = malloc(n_lines * sizeof(int));
c.day = malloc(n_lines * sizeof(int));
c.year = malloc(n_lines * sizeof(int));
c.newvalue = malloc(n_lines * sizeof(double));
使用它们完成后，请不要忘记free()。
我认为您应该分配array结构的n数组。最好称为coin，或者可能称为coin_value(或某种Camel_case名称，例如Coin_Value(，但它是基本的输入类型。它的复杂性很复杂，因为您需要3个阵列(253(结构的数组，但是它比在结构中隐藏单独的阵列要干净得多，这尤其是因为它极大地简化了内存分配(一个每个文件(。确实，可以一次分配所有内存(因此立即释放(，但我没有这样做。该代码还没有检查数据文件是否具有相同数量的行 - 应该。也没有检查相应条目中的日期值是否匹配；同样，可能应该进行这样的检查。当发现错误时，它将其报告在标准错误并退出。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct Coin_Value
{
int month;
int day;
int year;
double value;
} Coin_Value;
int main(int argc, char *argv[])
{
FILE *csv;
long n_lines = 0;
int n = argc - 1;
Coin_Value **m = (Coin_Value **)malloc(n * sizeof(*m));
if (m == 0)
{
fprintf(stderr, "Memory allocation failure (%zu bytes)n", n * sizeof(*m));
exit(1);
}
/* Should check that number of lines in each file is consistent */
for (int z = 1; z < argc; z++)
{
if ((csv = fopen(argv[z], "r")) == NULL)
{
fprintf(stderr, "%s not foundn", argv[z]);
exit(1);
}
if (fscanf(csv, "%li", &n_lines) != 1)
{
fprintf(stderr, "failed to read a number from %sn", argv[z]);
exit(1);
}
if (n_lines <= 0 || n_lines > 1000)
{
fprintf(stderr, "number of lines in %s out of control (got %ld)n", argv[z], n_lines);
exit(1);
}
/* Gobble any trailing data and newline */
int c;
while ((c = getc(csv)) != EOF && c != 'n')
;
Coin_Value *v = (Coin_Value *)malloc(n_lines * sizeof(*v));
if (v == 0)
{
fprintf(stderr, "Memory allocation failure (%zu bytes)n", n_lines * sizeof(*v));
exit(1);
}
char line[256];
for (int i = 0; fgets(line, sizeof(line), csv) != NULL && i < n_lines; i++)
{
if (sscanf(line, "%d/%d/%d,%lf", &v[i].month, &v[i].day, &v[i].year, &v[i].value) != 4)
{
fprintf(stderr, "Format error processing line: %s", line);
exit(1);
}
}
m[z-1] = v;
fclose(csv);
}
/* Multi-column output */
putchar('n');
for (int z = 1; z < argc; z++)
printf("%s%19.3s", (z == 1) ? "" : " ", argv[z]);
putchar('n');
for (long i = 0; i < n_lines; i++)
{
for (int z = 1; z < argc; z++)
{
printf("%s%.2d/%.2d/%.2d %9.2f", (z == 1) ? "" : " ",
m[z - 1][i].month, m[z - 1][i].day, m[z - 1][i].year, m[z - 1][i].value);
}
putchar('n');
}
putchar('n');
for (int z = 1; z < argc; z++)
{
printf("%.3s:n", argv[z]);
for (long i = 0; i < n_lines; i++)
{
printf("%.2d/%.2d/%.2d %9.2fn",
m[z - 1][i].month, m[z - 1][i].day, m[z - 1][i].year, m[z - 1][i].value);
}
putchar('n');
}
for (int z = 1; z < argc; z++)
free(m[z-1]);
free(m);
return 0;
}
此代码将数据打印两种不同的方式：
- 在页面上的n列中。
- 在页面下的n组条目中。
在您的样品输入上(5行，而不是253行(：
BTC NEO IOT
02/20/18 11403.70 02/20/18 128.36 02/20/18 1.91
02/19/18 11225.30 02/19/18 137.47 02/19/18 2.09
02/18/18 10551.80 02/18/18 127.38 02/18/18 1.98
02/17/18 11112.70 02/17/18 136.75 02/17/18 2.20
02/16/18 10233.90 02/16/18 128.85 02/16/18 2.10
BTC:
02/20/18 11403.70
02/19/18 11225.30
02/18/18 10551.80
02/17/18 11112.70
02/16/18 10233.90
NEO:
02/20/18 128.36
02/19/18 137.47
02/18/18 127.38
02/17/18 136.75
02/16/18 128.85
IOT:
02/20/18 1.91
02/19/18 2.09
02/18/18 1.98
02/17/18 2.20
02/16/18 2.10