错误:在初始化时无法将'std :: string'转换为'char *'

时间:2011-10-04 21:42:50

标签: c++ libcurl

我有从网页下载汇率的python脚本,我想从那里制作c ++程序,这是我到目前为止:

include iostream
include time.h
include stdio.h
include curl/curl.h
include curl/easy.h
include string
define CURL_STATICLIB

using namespace std;

void dat(string &d){
    time_t rawtime;
    struct tm * timeinfo;
    char datum[80];
    time ( &rawtime );
    timeinfo=localtime(&rawtime);
    strftime(datum,80,"%d%m%y",timeinfo);
    d=datum;
}

size_t write_data(void *ptr, size_t size, size_t nmemb, FILE *stream) {
    size_t written;
    written = fwrite(ptr, size, nmemb, stream);
    return written;
}


int main()
{
    string f;
    dat(f);
    string l1="http://www.hnb.hr/tecajn/f";
    string l2=".dat";
    string linkz=l1+f+l2;
    cout << linkz;

    CURL *curl;
    FILE *fp;
    CURLcode res;
    char *url = linkz;
    char outfilename[FILENAME_MAX] = "/home/tomi/data.txt";
    curl = curl_easy_init();
    if (curl) {
        fp = fopen(outfilename,"wb");
        curl_easy_setopt(curl, CURLOPT_URL, url);
        curl_easy_setopt(curl, CURLOPT_WRITEFUNCTION, write_data);
        curl_easy_setopt(curl, CURLOPT_WRITEDATA, fp);
        res = curl_easy_perform(curl);
        curl_easy_cleanup(curl);
        fclose(fp);


    return 0;
}

当我尝试编译时,它给我这个错误,我发现使用algorythm下载txt,所以我希望它是正确的

3 个答案:

答案 0 :(得分:6)

如果您已指出出现错误的行,我就不必将其追踪到:

string linkz=l1+f+l2;
...
char *url = linkz;

您可以使用c_str()获取指向字符串中const字符的指针。所以这样做:

char const* url = linkz.c_str();

您可以在setopt来电中使用同一行,也可以将url设为std::string

答案 1 :(得分:6)

char *url = linkz;应为const char* url = linkz.c_str();,假设您出于API原因确实需要C风格的字符串。

答案 2 :(得分:3)

问题出在这一行:

char *url = linkz;

“links”是一个std :: string,但“url”是一个char *。尝试使用string的c_str方法来获得所需的内容:

const char * url = links.c_str();