argc

C++: cin >> *char

浪尽此生 提交于 2020-04-13 17:23:35
问题 So, I'm currently writing a line editor as a learning project on I/O, writing files, and the like. It is written in C++, and I am currently trying to write out to a file of the user's choosing. I have CLI arguments implemented, but I currently have no idea how to implement an in program way of specifying the file to write to. char *filename; if (argc >= 2){ filename = argv[1]; } else{ cout << "file>"; cin >> filename; cin.ignore(); } This works perfectly well when I use command line arguments

replacing the command line arguments int argc and char** argv with std::vector<std::string>

三世轮回 提交于 2020-03-14 18:37:48
问题 Following this post, where I have found a temporary workaround for my other problem, I want to know if I can replace the int argc, char** argv with a std::vector<std::string> variable/object. Consider the imaginary code: #include <iostream> #include <CloseLibrary> void someFunction(int argc, char** argv){ for (int i = 0; i < argc; ++i) { std::cout << argv[i] << std::endl; } } int myFunc(int argc, char** argv){ someFunction(argc, argv); return 0; } where the CloseLibrary is a closed library

replacing the command line arguments int argc and char** argv with std::vector<std::string>

我的梦境 提交于 2020-03-14 18:37:20
问题 Following this post, where I have found a temporary workaround for my other problem, I want to know if I can replace the int argc, char** argv with a std::vector<std::string> variable/object. Consider the imaginary code: #include <iostream> #include <CloseLibrary> void someFunction(int argc, char** argv){ for (int i = 0; i < argc; ++i) { std::cout << argv[i] << std::endl; } } int myFunc(int argc, char** argv){ someFunction(argc, argv); return 0; } where the CloseLibrary is a closed library

int main(int argc, char *argv[])主函数参数

岁酱吖の 提交于 2019-12-27 18:09:26
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> int main(int argc, char *argv[]) int main(int argc, char **argv) argc: 传入参数的个数 argv:传入的参数 argv[0]:指向程序运行时的全路径名称 argv[1]: 第一个参数 shell脚本里的传参也是这样的 $1 $2 表示 第一个参数 和第二个参数,在运行程序时把参数添加上 参考: 1、 http://wiki.opencv.org.cn/index.php/Main函数参数argc,argv说明 2、还有一篇写的清晰还带截图的链接找不到了,百度应该有 来源: oschina 链接: https://my.oschina.net/u/1240964/blog/655809

How to exclude arguments passed from command prompt argc argv in C? [duplicate]

可紊 提交于 2019-12-25 02:43:30
问题 This question already has answers here : Reading fractions in C (6 answers) Closed 5 years ago . I need to add fractions given by the user through command prompt in the format a/b b/c I thought I could do it this way: n1 = atoi(argv[1]); d1 = atoi(argv[3]); n2 = atoi(argv[4]); d2 = atoi(argv[6]); Thereby skipping the slashes, but this just crashes the program. Is there some way to skip over certain characters passed as arguments through command prompt? Thank you in advance. 回答1: If the user

Jython 2.5.1: Calling From Java into __main__ - how to pass in command line args?

心不动则不痛 提交于 2019-12-21 17:52:06
问题 I'm using Jython from within Java; so I have a Java setup similar to below: String scriptname="com/blah/myscript.py" PythonInterpreter interpreter = new PythonInterpreter(null, new PySystemState()); InputStream is = this.getClass().getClassLoader().getResourceAsStream(scriptname); interpreter.execfile(is); And this will (for instance) run the script below: # myscript.py: import sys if __name__=="__main__": print "hello" print sys.argv How I pass in 'commandline' arguments using this method ?

Passing arguments to executable from command line

风流意气都作罢 提交于 2019-12-14 02:15:42
问题 I'm trying to pass arguments to a Fortran executable from the command line. A sample program that achieves this in C is (taken from here): #include <stdio.h> int main (int argc, char *argv[]) { int count; printf ("This program was called with \"%s\".\n",argv[0]); if (argc > 1) { for (count = 1; count < argc; count++) { printf("argv[%d] = %s\n", count, argv[count]); } } else { printf("The command had no other arguments.\n"); } return 0; } The output of this program is: This program was called

Arguments in main() ignored when debugging in Visual C++

限于喜欢 提交于 2019-12-14 01:22:42
问题 Up to this point I've been able to correctly view the output of my C codes by using the debugging command in Visual C++. However, when the script relies on parameters in the main function (eg/ argc , argv ), the debugger seems to ignore both parameters and treat them as though they are uninitialized. For instance, in the following code, the output is always printf("Usage: find pattern\n"); . #include <stdio.h> #include <string.h> #define MAXLINE 1000 int getline(char *line, int max); /* find:

Checking to make sure argv[1] is an integer c++

断了今生、忘了曾经 提交于 2019-12-13 16:47:05
问题 For my program I have to make sure the user only inputs a positive INTEGER. for example if the user inputted 12hi it should not run the program and print to std error. I am not quite sure how to implement this. int main(int argc, char *argv[]) { if(atoi(argv[1]) < 1) { cerr << "ERROR!"<< endl; return 1; } return 0; } 回答1: Pass it to a std::istringstream and ensure all data was processed: if (a_argc > 1) { std::istringstream in(a_argv[1]); int i; if (in >> i && in.eof()) { std::cout << "Valid

OpenCV argc and argv confusion

人走茶凉 提交于 2019-12-11 01:37:20
问题 I'm checking some OpenCV tutorial and found this line at the beginning (here is the link, code is under the CalcHist section http://opencv.willowgarage.com/documentation/c/histograms.html) if (argc == 2 && (src = cvLoadImage(argv[1], 1)) != 0) I've never seen this before and really don't understand it. I checked some Q&A regarding this subject but still don't understand it. Could someone explain to me what is the meaning of this line? Thanks! 回答1: The line does the following, in order: Tests