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下面是我的代码：快速排序的结构C
我似乎无法有效地使用快速排序......事实证明我的数组0的它们与名称填充和开始时间后......它是一个问题我qsort电话？或qsort本身。
头与结构如下：
/**
* Simulation of a process scheduler
*/
//#ifndef SCHEDULER_H_
#define SCHEDULER_H_
#include <stddef.h>
#include <errno.h>
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <ctype.h>
/* types */
/** units of time */
typedef long time;
/** process identifier */
typedef int pid;
/** Information about a job of interest to the task scheduler */
struct job_data {
/* pid of this process */
pid pid;
/* time process starts */
time start;
/* time needed to finish */
time finish;
/* time spent processing so far */
time scheduled;
/* Number of lines */
int lines;
};
struct job {
/* Various parameters used by the scheduler */
char job_id[20];
struct job_data parameters;
char *lines[20];
};
/* I/O Files */
//static char *inputFile;
char * in;
static FILE *input;
static FILE *cur;
/*Scheduled jobs indexed by PID*/
struct job list[20];
/* the next job to schedule */
//static struct job *job_next = NULL;
/* Time */
time clock;
/*Comparison for qsort*/
int compare_start(const void *x, const void *y)
{
const struct job *a = x;
const struct job *b = y;
printf("%ld, %ld\n", a->parameters.start, b->parameters.start);
if (a->parameters.start < b->parameters.start)
{
return -1;
}
if (a->parameters.start > b->parameters.start)
{
return 1;
}
return 0;
}
/*Order Jobs*/
static void order_jobs(void)
{
qsort(list, (sizeof list)/(sizeof list[0]), sizeof list[0], compare_start);
}
/** Read and parse input from input file */
static void parse_input(void)
{
char buffer[BUFSIZ];
char lines[BUFSIZ];
int jobs = 0;
struct job *current;
while(fgets(buffer, sizeof(buffer), input))
{
time start;
char buf[BUFSIZ];
sscanf(buffer,"./%s/", buf);
cur = fopen(buf, "r");
int n_lines = 0;
while(fgets(lines, sizeof(lines), cur))
{
if(n_lines == 0)
{
current = &list[jobs];
strcpy(current->job_id, buf);
sscanf(lines,"%ld", &start);
current->parameters.start = start;
}
n_lines++;
}
current->parameters.lines = n_lines;
jobs++;
fclose(cur);
}
order_jobs();
for (int i = 0; i < jobs; i++)
{
printf("%s %ld %d\n", list[i].job_id, list[i].parameters.start, list[i].parameters.lines);
}
}
int main(int argc, char **argv)
{
in = argv[1];
if ((input = fopen(in, "r")) == NULL) {
fprintf(stderr, "cannot open %s\n", argv[1]);
}
parse_input();
fclose(input);
return EXIT_SUCCESS;
}
您应该只发布代码的相关部分。试着更具体地说明你的问题。 – Constantinius
有关@Constantinius的含义的更详细解释，请阅读本网站关于[简短，独立，正确（可编译），示例]（http://sscce.org/）。例如：你的例子依赖于复杂的结构，你是否可以用一个或两个字段的结构演示类似的问题？你可以写一个不能做I/O的版本，但是仍然可以证明这个问题？你无论如何都得到了答案，但是*对未来来说，遵循这些方法是一个好主意，你可以在这个过程中自己找到自己的吸烟枪...... – HostileFork