Question

我有一个大约2500行的数据文件。每行包含5个参数。我宣布了一个常规方面的支持。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<sys/wait.h>
#include<stdbool.h>

char *copy_string(const char *str) {
    char *s = malloc(strlen(str) + 1);
    if (s) strcpy(s, str); else {perror("malloc"); exit(1);}
    return s;
}

void func(char **arr, char *w)
{
    int i = 0, j = 0, k = 0;

    char temp[100];

    for (i = 0; i < 100; i++)
    {
        if (w[i] == '\0' || w[i] == '\n')
        {
            temp[j] = '\0';
            arr[k] = copy_string(temp);
            arr[k+1] = NULL;
            break;
        }
        if (w[i] == ' ')
        {
            temp[j] = '\0';
            arr[k] = copy_string(temp);
            k++;
            j = 0;
        }
        else
        {
            temp[j] = w[i];
            j++;
        }

    }
}
int main(void)
{
    char *n = "/bin/ls";
    char *arr[10] = {NULL};
    char p[100] = {0};
    char *w = "ls -l";
    int i = 0;
    //printf("bilal-hassan-qadri $ >>");
    fgets(p, 100, stdin);
    w = p;
    arr[2] = NULL;
    bool found = false;
    for (i = 0; w[i] != '\0'; i++)
    {
        if (w[i] == ' ')
        {
            found=true;
            func(arr,w);
            break;
        }
    }
    if (!found)
      arr[0] = w;
    int status;
    int id = fork();
    if (id == 0)
    {
        if (execv(n,arr) < 0)
        {
            printf("invalid commandn");
        }
        else
        {
            printf("ninvalid command");
        }
    }
    else
    {
        wait(&status);
        for (i = 0; arr[i] != NULL; i++) free(arr[i]);
    }
    return 0;
}

我留出内存将文件的每一行读入一个结构数组。

typedef struct data
{

char date[9];
char time[3];
float price;
float NAV;
float percent_change;
} _data ;

void readfile(int *j, struct data *input);

void readfile(int *j,struct data *input)

并阅读文件。

input = calloc(days_of_data,sizeof(*input));
for (i = 0; i< days_of_data; i++)
    input[i] = *(_data*)calloc(1,sizeof(_data));

这很有效。现在我想将结构发送到主程序。

while(fgets(line, 75, fp) != NULL)
{      

    date = strtok(line,",");
    strcpy(input[i].date,date);
    time = strtok(NULL,",");
    strcpy(input[i].time, time);
    price = strtok(NULL,",");
    input[i].price = atof(price);
    NAV = strtok(NULL,",");
    input[i].NAV = atof(NAV);
    percent_change = strtok(NULL,",");
    input[i].percent_change =  atof(percent_change);
    i--;

}

这会在print语句中编译但崩溃。

Answer 1

您必须使用返回值

void readfile(int *j, struct data **input_out) {
    struct data *input;
    /* read the data */
    *input_out = input;
}

int main(int argc, const char * argv[]) {
    struct data *input;
    readfile(&j, &input);
    printf("%f\n",input[0].percent_change);
}

或指针

package main

import (
    "fmt"
    "reflect"
)

type TT struct {
    Foo int
}

type II interface {
    Bar(int) (int, error)
}

type SS struct {
    F1 TT
    F2 II
}

func main() {
    var rr SS
    value := reflect.ValueOf(rr)
    for ii := 0; ii < value.NumField(); ii++ {
        fv := value.Field(ii)
        xv := fv.Interface()
        switch vv := xv.(type) {
        default:
            fmt.Printf("??: vv=%T,%v\n", vv, vv)
        case TT:
            fmt.Printf("TT: vv=%T,%v\n", vv, vv)
        case II:
            fmt.Printf("II: vv=%T,%v\n", vv, vv)
        }
    }
}

将数据传递给调用者。

将一个结构数组从一个函数传递给主程序

1 个答案: