argv

For some reason my C program is refusing to convert elements of argv into ints, and I can't figure out why. int main(int argc, char *argv[]) { fprintf(stdout, "%s\n", argv[1]); //Make conversions to int int bufferquesize = (int)argv[1] - '0'; fprintf(stdout, "%d\n", bufferquesize); } And this is the output when running ./test 50: 50 -1076276207 I have tried removing the (int), throwing both a * and an & between (int) and argv[1] - the former gave me a 5 but not 50, but the latter gave me an output similar to the one above. Removing the - '0' operation doesn't help much. I also tried making a

First problem: Suppose we write a simple program which takes in command line arguments and prints to a file. If the user enters writetofile Hello!0\n w%orl\t!@#y bash replies with !0: event not found. Without the user knowing things like using quotes ('') or escape characters ('\'), how do I handle this stuff instead of bash understanding it as a command? Second problem: Once I get these arguments, how do I interpret them as special characters and not sequences of characters. (ie. \t is tab, not '\''t') That is, how do make sure that the program writes this to the file: Hello!0 w%orl !@#y and

I have the following code which reads an file name from the command line and opens this file: #include <stdio.h> #include <stdlib.h> int main(int argc, char **argv){ FILE *datei; char filename[255]; //filename = argv[1]; //datei=fopen(filename, "r"); datei=fopen(argv[1], "r"); if(datei != NULL) printf("File opened"); else{ printf("Fehler beim öffnen von %s\n", filename); return EXIT_FAILURE; } return EXIT_SUCCESS; } This example works, but I want to write the string from the command line to the char array and pass that char array to to fopen(), but i get the compiler error Error: assignment to

I need to get argv[1] and argv[2] to different types. I found that I could only use sscanf() once or the next string in argv cannot be retrieved. Here's my code. int main( int argc, char *argv[]) { char t; float temp; sscanf(argv[1], "-%[cf]",&t); sscanf(argv[2], "%f", &temp); return 0; } Only the first sscanf() can get the formatted value. How could I also get done with argv[2]? Attempt to save string data in a char leading to undefined behavior (UB). "%[]" expects to match a character array. // char t; // sscanf(argv[1], "-%[cf]",&t); char t[100]; if (sscanf(argv[1], "-%99[cf]",t) != 1)

I'm using a shared host (fasthostingdirect) and for some reason they have this flag turned off by default. This means I'm unable to access PHP command line parameters... unless I use the -n (= --no-php-info ) flag after php.exe . Have tried ini_set('register_argc_argv', 1) in my php file but it has no effect. Am guessing this is due to the clamped down nature of the hosting provider, however they don't stop the -n option - not sure of the other implications of using this though. Does anyone have any better suggestions? The ini_set('register_argc_argv', 1) does not work because by the time the

This question already has an answer here: Is argv[n] writable? 4 answers Recently (Jan 2016, in case the question persists long enough) we had the question Are the strings in argv modifiable? . In the comment section to this answer, we (@2501 and I) argued whether it is really the strings of characters (an example character being **argv ) that's modifiable or the pointers to the strings (an example pointer being *argv ). The appropriate standard quotation is from the C11 standard draft N1570, §5.1.2.2.1/2: The parameters argc and argv and the strings pointed to by the argv array shall be