求指点 字符串转换时间戳 strncpy分割字符，移动原字符指针后，复制无效了

four498

#include <stdio.h>
#include <iostream>
#include <sys/time.h>
using namespace std;

int str_to_time(char *sdate){
    std::cout<<"aaa:"<< sdate <<std::endl;//打印需要转换时间戳字符
    time_t t = time(NULL);
    struct tm *betime;
    char year[4],mm[2],dd[2],h[2],i[2],s[2];
    strncpy(year,sdate,4);//截取4个字符
    year[4]='\0';
    printf("year:%s\n",year);//第一次打印
    if(strcmp(year,"0000")==0) return 0;
    strncpy(mm,sdate+5,2);//截取月
    mm[2]='\0';
    printf("mm:%s\n",mm);//第一次打印
    strncpy(dd,sdate+8,2);
    dd[2]='\0';
    printf("dd:%s\n",dd);
    strncpy(h,sdate+11,2);
    h[2]='\0';
    printf("h:%s\n",h);
    strncpy(i,sdate+14,2);
    i[2]='\0';
    printf("i:%s\n",i);
    strncpy(s,sdate+17,2);
    s[2]='\0';
    printf("s:%s\n",s);

    printf("mm:%s\n",mm);//第二次打印，为空了
    printf("year:%s\n",year);//第二次打印，为空了
    betime->tm_year = atol(year)-1900;
    betime->tm_mon = atol(mm);
    betime->tm_mday = atol(dd);
    betime->tm_hour = atol(h);
    betime->tm_min = atol(i);
    betime->tm_sec = atol(s);

    t = timegm(betime);
    int a = time(&t);
    return a;
};

int main(int argc,char *argv[]){
    char sdate[20] = "2010-12-11 17:23:24";
    int da = str_to_time(sdate);
    cout << sdate << ":::" << da << endl;

    return 0;

};

编译后运行结果:
aaa:2010-12-11 17:23:24
year:2010
mm:12
dd:11
h:17
i:23
s:24
mm:
year:
Segmentation fault

为什么 mm year的值会没有了？

four498

继续等待

hellioncu

char数组的定义长度都加1

four498

加一运行结果：

aaa:2010-12-11 17:23:24
year:2010
mm:12
dd:11
h:17
i:23
s:24
mm:12
year:2010
i:23
a=1288704162
Segmentation fault


好像可以了，为什么了？
为什么还出现Segmentation fault？这里返回数字了。

four498

难道是strncpy(s,sdate+17,2);后越位了？

xautLL

struct tm *betime; 声明了没有分配空间就用了，造成溢出了。但是运气不佳，没有在使用betime的时候就出段错误，在函数返回的时候出错了

four498

回复 6# xautLL


是的，gdb调试总显示cout<<时候内存错误。疑惑了一个晚上。

four498

最后给出完整函数，下面再给一个实现相同功能的方法，总结于此：
#include <stdio.h>
#include <iostream>
#include <sys/time.h>
#include <string.h>
using namespace std;

int str_to_time(char *sdate){
    printf("sdate:%s\n",sdate);
    time_t t = time(NULL);
    struct tm *betime = (struct tm*)malloc(sizeof(struct tm));//申请时间结构体空间
    char year[5],mm[3],dd[3],h[3],i[3],s[3];//注意字符大少，
    strncpy(year,sdate,4);
    year[4]='\0';//加入结束字符
    printf("year:%s\n",year);//第一次输出测试
    if(strcmp(year,"0000")==0) return 0;
    strncpy(mm,sdate+5,2);
    mm[2]='\0';
    printf("mm:%s\n",mm);
    strncpy(dd,sdate+8,2);
    dd[2]='\0';
    printf("dd:%s\n",dd);
    strncpy(h,sdate+11,2);
    h[2]='\0';
    printf("h:%s\n",h);
    strncpy(i,sdate+14,2);
    i[2]='\0';
    printf("i:%s\n",i);
    strncpy(s,sdate+17,2);
    s[2]='\0';
    printf("s:%s\n",s);

    printf("mm:%s\n",mm);
    printf("year:%s\n",year);//第二次打印测试

    betime->tm_year = (int) (atol(year)-1900);//gdb时候减法后是long，这里转换为int
    betime->tm_mon = atol(mm);
    betime->tm_mday = atol(dd);
    betime->tm_hour = atol(h);
    betime->tm_min = atol(i);
    betime->tm_sec = atol(s);
    printf("i:%s\n",i);

    t = timegm(betime);
    int a = time(&t);
    printf("a=%d\n",a);//测试输出
    free(betime);//释放tm结构体内存
    return a;
};

int main(int argc,char *argv[]){
    char sdate[22];
    snprintf(sdate,sizeof(sdate),"%s","2010-12-11 17:23:24")
    //cout << sdate << endl;
    int da = str_to_time(sdate);
    printf("da=%d\n",da);
    return 0;
}

four498

另一实现：
#include <stdio.h>
#include <iostream>
#include <sys/time.h>
#include <string.h>
using namespace std;

int str_to_time(char *sdate){
    struct tm* tmp_time = (struct tm*)malloc(sizeof(struct tm));

    strptime(sdate,"%Y-%m-%d %H:%M:%S",tmp_time);
    time_t t = mktime(tmp_time);
    cout << t << endl;
        free(tmp_time);
        return t;
};

int main(int argc,char *argv[]){
    char sdate[22];
    snprintf(sdate,sizeof(sdate),"%s","2010-12-11 17:23:24");

    int da = str_to_time(sdate);
    printf("da=%d\n",da);
    return 0;
}