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POHow can I copy the contents of argv[] into a c style string?
 

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  1. POHow can I copy the contents of argv[] into a c style string?
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    <p>I'm trying to write a program that takes several arguments at runtime to append text to a file. </p> <p>The program produces a segmentation fault at runtime. Here is the code: </p> <pre><code>int main(int argc, char* argv[]) { //error checking if (argc &lt; 1 || argc &gt; 4) { cout &lt;&lt; "Usage: -c(optional - clear file contents) &lt;Filename&gt;, message to write" &lt;&lt; endl; exit(EXIT_FAILURE); } char* filename[64]; char* message[256]; //set variables to command arguments depending if -c option is specificed if (argc == 4) { strcpy(*filename, argv[2]); strcpy(*message, argv[3]); } else { strcpy(*filename, argv[1]); strcpy(*message, argv[2]); } int fd; //file descriptor fd = open(*filename, O_RDWR | O_CREAT, 00000); //open file if it doesn't exist then create one fchmod(fd, 00000); return 0; } </code></pre> <p>I am still quite a beginner and I'm having immense trouble understanding c strings. What's the difference between char* and char[] and char* []? </p> <p>UPDATE:</p> <p>The code still throws a segmentation fault, here is my revised code:</p> <pre><code>using std::cout; using std::endl; int main(int argc, char* argv[]) { //error checking if (argc &lt; 1 || argc &gt; 4) { cout &lt;&lt; "Usage: -c(optional - clear file contents) &lt;Filename&gt;, message to write" &lt;&lt; endl; exit(EXIT_FAILURE); } char filename[64]; char message[256]; //set variables to command arguments depending if -c option is specificed if (argc == 4) { strncpy(filename, argv[2], 64); strncpy(message, argv[3], 256); } else { strncpy(filename, argv[1], 64); strncpy(message, argv[2], 256); } int fd; //file descriptor fd = open(filename, O_RDWR | O_CREAT, 00000); //open file if it doesn't exist then create one fchmod(fd, 00000); return 0; } </code></pre>
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    1. COYou get points for providing a complete attempt at the program; plenty to learn, but you were close.
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