C assign string from argv[] to char array

问题

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 expression with array type filename = argv[1];

What does this error mean and what can I do to fix it?

回答1:

You must copy the string into the char array, this cannot be done with a simple assignment.

The simplistic answer is strcpy(filename, argv[1]);.

There is a big problem with this method: the command line parameter might be longer than the filename array, leading to a buffer overflow.

The correct answer therefore:

if (argc < 2) {
    printf("missing filename\n");
    exit(1);
}
if (strlen(argv[1]) >= sizeof(filename)) {
    printf("filename too long: %s\n", argv[1]);
    exit(1);
}
strcpy(filename, argv[1]);
...

You might want to output the error messages to stderr. As a side note, you probably want to choose English or German, but not use both at the same time ;-)

来源：https://stackoverflow.com/questions/28680557/c-assign-string-from-argv-to-char-array