我有这段代码通过libcurl发送电子邮件
#define FROM "<my@gmail.com>"
#define TO "<your@gmail.com>"
static const char *payload_text[] = {
"To: " TO "\r\n",
"From: " FROM "\r\n",
"Subject: Shift\r\n",
"text\r\n",
NULL
};
struct upload_status {
int lines_read;
};
static size_t payload_source(void *ptr, size_t size, size_t nmemb, void *userp)
{
struct upload_status *upload_ctx = (struct upload_status *)userp;
const char *data;
if((size == 0) || (nmemb == 0) || ((size*nmemb) < 1)) {
return 0;
}
data = payload_text[upload_ctx->lines_read];
if(data) {
size_t len = strlen(data);
memcpy(ptr, data, len);
upload_ctx->lines_read++;
return len;
}
return 0;
}
int main(void)
{
CURL *curl;
CURLcode res = CURLE_OK;
struct curl_slist *recipients = NULL;
struct upload_status upload_ctx;
upload_ctx.lines_read = 0;
curl = curl_easy_init();
if(curl) {
curl_easy_setopt(curl, CURLOPT_USERNAME, "my@gmail.com");
curl_easy_setopt(curl, CURLOPT_PASSWORD, "pass");
curl_easy_setopt(curl, CURLOPT_URL, "smtp://smtp.gmail.com:587");
curl_easy_setopt(curl, CURLOPT_USE_SSL, (long)CURLUSESSL_ALL);
curl_easy_setopt(curl, CURLOPT_CAINFO, "google.pem");
curl_easy_setopt(curl, CURLOPT_MAIL_FROM, FROM);
recipients = curl_slist_append(recipients, TO);
//curl_easy_setopt(curl, CURLOPT_FILE, "edgE0DF.tmp");
curl_easy_setopt(curl, CURLOPT_MAIL_RCPT, recipients);
curl_easy_setopt(curl, CURLOPT_READFUNCTION, payload_source);
curl_easy_setopt(curl, CURLOPT_READDATA, &upload_ctx);
curl_easy_setopt(curl, CURLOPT_UPLOAD, 1L);
curl_easy_setopt(curl, CURLOPT_VERBOSE, 1L);
res = curl_easy_perform(curl);
if(res != CURLE_OK)
fprintf(stderr, "curl_easy_perform() failed: %s\n",
curl_easy_strerror(res));
curl_slist_free_all(recipients);
curl_easy_cleanup(curl);
}
return 0;
}
我的问题是:如何创建一个函数,该函数将来自和来自的参数作为参数并更改我的全局FROM
和TO
?这可能吗?因为,我知道全局变量在代码的其他部分之前被初始化,这意味着使用static const char *payload_text[]
和FROM
的{{1}}将使用错误的地址进行初始化。