argv

问题 I recently came across the following in my searches regarding environment variables in C: int main (int argc, char *argv[], *char *envp[]) I have searched around and can't find anything conclusive regarding my question. What are all of the available arguments that main() can accept? 回答1: The C99 and C11 draft standards allow for implementation defined set of parameters to main , these parameters are going to be specific to those systems( non-portable ). From section 5.1.2.2.1 : [...]or in

问题 I want to learn more C++... Usually I make a for loop to parse argv, and I wind up with a bunch a C-style strings. I want to do something similar in C++, but preferably without reading from /proc/whatever. At first, I tried to convert the C-style string to a C++ style string without results... The frustrating bit is that everyone on SO seems to want to know how to go the other way, which is what c_str() is for. What's a good C++ way to do this (ie parse argv)? Also, one note, I'm looking for

问题 My main function is as follows: int main(int argc, char const *argv[]) { huffenc(argv[1]); return 0; } The compiler returns the warning: huffenc.c:76: warning: passing argument 1 of ‘huffenc’ discards qualifiers from pointer target type For reference, huffenc takes a char* input, and the function is executed, with the sample input "senselessness" via ./huffenc senselessness What could this warning mean? 回答1: It means that you're passing a const argument to a function which takes a non- const

问题 Errors while calling int getopt function from http://code.google.com/p/darungrim/source/browse/trunk/ExtLib/XGetopt.cpp?r=17 `check.cpp: In function ‘int main()’:` check.cpp:14:55: error: invalid conversion from ‘const char**’ to ‘char* const*’ [-fpermissive] /usr/include/getopt.h:152:12: error: initializing argument 2 of ‘int getopt(int, char* const*, const char*)’ [-fpermissive] #include <iostream> #include <cstring> #include <string> #ifdef USE_UNISTD #include <unistd.h> #else #include

问题 I pass an executable on the command-line to my python script. I do some calculations and then I'd like to send the result of these calculations on STDIN to the executable. When it has finished I would like to get the executable's result back from STDOUT. ciphertext = str(hex(C1)) exe = popen([sys.argv[1]], stdout=PIPE, stdin=PIPE) result = exe.communicate(input=ciphertext)[0] print(result) When I print result I get nothing, not None, an empty line. I'm sure that the executable works with the

问题 I'm trying to use fscanf to read and print every character on the screen, but I'm getting a segmentation fault (core dumped) when I run the program. Here's my code: #include <stdio.h> main(int argc, char * argv[]) { int *a ; FILE *input; if (argc>=2) { input= fopen(argv[1],"r"); if (input!=NULL) { while (feof(input)==0) { fscanf(input,"%d\n",a); printf("%d\n",*a); } fclose(input); } else { printf("Error!\n"); } } } I provide the file as an argument, like this: ./myprog input.txt The file

问题 What is the standard practice in Python when I have a command-line application taking one argument which is URL to a web page or path to a HTML file somewhere on disk (only one) is sufficient the code? if "http://" in sys.argv[1]: print "URL" else: print "path to file" 回答1: Depends on what the program must do. If it just prints whether it got a URL, sys.argv[1].startswith('http://') might do. If you must actually use the URL for something useful, do from urllib2 import urlopen try: f =

问题 #include <stdio.h> int main(int argc, char *argv[]) { int i; for(i=1;i<argc;i++) printf("%s%s", argv[i], (i<argc-1)? " ":""); printf("\n"); return 0; } Given above is a simple C program that outputs command line inputs. Here argc is the argument counter. argv is said to be an array that contains arguments. My question is: why does it define as a pointer to a character array instead of a normal array? Also what is the need for defining its zeroth element (argv[0]) as the name by which the

问题 I noticed that the main.cpp in a Qt application has to contain the following line: QApplication app(argc, argv); I know that argc is the number of command-line arguments, and argv is that array list of command-line arguments. But, the question in my mind is: what are those arguments I'm passing to the constructor and at the same time cannot explicitly see? What is working behind the scences out there? Thanks. 回答1: There are no hidden arguments. You can explicitly see every argument- argc,

问题 How do I find a program's argc and argv from a shared object? I am writing a library in C that will be loaded via LD_PRELOAD . I've been able to find the stack two different ways: Read rsp via inline __asm__ call. Read /proc/<pid>/maps and parse the entry for stack. I can then create a pointer, point it at the stack segment, then iterate through looking for data. The problem is I can't figure out an efficient way to determine what bytes are argc and the pointer to the pointer to the argv