argv

How can I input text into TCHAR* argv[] ? OR: How can I convert from char to TCHAR* argv[] ? char randcount[] = "Hello world"; TCHAR* argv[]; argv = convert(randcount); One way to do is: char a[] = "Hello world"; USES_CONVERSION; TCHAR* b = A2T(a); /*This code did TCHAR in my project without A2T or any other converters. Char text is a some kind of array. So we can take letters one by one and put them to TCHAR. */ #include <iostream> TCHAR* Converter(char* cha) { int aa = strlen(cha); TCHAR* tmp = new TCHAR[aa+1]; for(int i = 0; i< aa+1; i++) { tmp[i]=cha[i]; } return tmp; } int main() { char*

I'm new to C and I'm thinking how to write this function myself. I take a parameter from command line, so it is stored in argv array and I want to decide whether it is or isn't number. What is the easiest way to do this? Thank you #include <stdio.h> int isNumber(int *param) { if (*param > 0 && *param < 128) return 1; return 0; } int main(int argc, char *argv[]) { if (argc == 2) isNumber(argv[1]); else printf("Not enought parameters."); return 0; } Read about strtol(3) . You could use it as bool isnumber(const char*s) { char* e = NULL; (void) strtol(s, &e, 0); return e != NULL && *e == (char)0;

Quoting from docs.python.org : " sys.argv The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c' . If no script name was passed to the Python interpreter, argv[0] is the empty string." Am I missing something, or sys.argv[0] always returns the script name, and to get '-c' I'd have to use sys.argv[1] ? I'm testing with Python 3.2 on GNU/Linux. No, if you invoke Python with -c

If I have this: int main(int argc, char *argv[]) In the body, you can sometimes find programs using argv[1] . When do we use argv[1] over argv[0] ? Is it only when we just want to read the second argument in the command line? By convention , argv[0] is the current program's name (or path), and argv[1] through argv[argc - 1] are the command-line arguments that the user provides. However, this doesn't have to be true -- programs can OS-specific functions to bypass this requirement, and this happens often enough that you should be aware of it. (I'm not sure if there's much you can do even if you

This question already has an answer here: argv[argc] ==? 2 answers I read an article (forgot the URL), which said that argv[argc] is a NULL pointer (contains \0 ). To check whether if its true I wrote this code, yeah it exist. What I don't understand is, why does the OS include this NULL pointer at argv[argc] . Is it useful for something else also? int main (int argc, char **argv){ while (*argv) printf ("%s\n", *argv++); return 0; } The C Standard 5.1.2.2.1/2 second mark says explicitly argv[argc] shall be a null pointer. The C++ Standard 3.6.1/2 also says explicitly The value of argv[argc]

#include <stdio.h> #include <stdlib.h> #include <string.h> int main(int argc, char *argv[]){ int fir; //badly named loop variable char *input[] = calloc( strlen(argv), sizeof(char)); //initializing an array for( fir = 1; fir< strlen(argv); fir++){ //removing the first element of argv strcat(input, argv[fir]); // appending to input } } The error I'm getting is for line 7. It says "passing argument 1 of 'strlen' from incompatible pointer type". I get the same error for the strcat function. It also says "given a char ** but expected a const char * " for both functions. I'm trying to populate a