它想知道如何创建 - 我的意思是C ++示例 - .exe文件中的.exe文件? 不想使用Launch4j或类似的。只是想从头开始。
答案 0 :(得分:1)
所以答案是here。它看起来像:
#include <iostream>
#include <fstream>
#include <windows.h>
using namespace std;
string getAppName() {
// Creates char array with maximum needed length
char result[MAX_PATH];
// Uses <windows.h> method to retrieve app name. Only Windows OS
std::string(result, GetModuleFileName(NULL, result, MAX_PATH));
char* token;
string name, next = "";
//Iterates until end of tokens of Name string, divided by char -> '\' and '.'
//The only '.' in Filename must be that for ".exe" , error otherwise.
token = strtok(result, "\\.");
while (next != "exe") {
name = next;
token = strtok(NULL, "\\.");
next.assign(token);
}
return name;
}
int main(int argc, char** argv) {
ifstream file; // file stream
string value = "";
string jar_name = "";
string image_name = "splash.gif";
file.open("run.ini");
if (!file) { // file does not exist, so use standard name
jar_name = getAppName();
jar_name += ".jar";
} else {
do {
file >> jar_name;
file >> image_name;
} while (!file.eof());
file.close();
}
//Launch JAVAW command on "jar_name" file with "image_name" splash image
string exec_command = "start javaw -splash:" + image_name + " -jar " + jar_name;
system(exec_command.c_str());
return 0;
}
答案 1 :(得分:0)
的#include void main(void) { system(&#34; Java&lt; Filename&gt;&#34;);
另存为呼叫中心 properties:compile-&gt; make as .exe 从tc / bin
获取exe文件