解析命令行参数时出现分段错误

Question

I am getting a segmentation fault I am having trouble isolating the source of the fault. I believe that for some reason argv and argc are not getting values, but I cannot figure out why.

Here is the first part of my code.

int main(int argc, char **argv)
{
int i;
int numArrays = 0;
int test = 0;
int opt = 0;

printf ("Parsing....\n");  // This line gets printed
printf(argv);    // this one doesn't even if being flushed immediately after
printf(argc);    // when the previous line is deleted, this line also won't print
fflush(stdout); 
opt =  getopt(argc, argv, optString);  //  defined as optString = "qmc";

while (opt != -1) {
    printf (opt);
    switch (opt) {
    //prevents program from printing to a file
    case 'q':
        tofile = 1;
        break;
    // max size of the prime numbers
    case 'm':
        maxPrime = atoi(optarg);
        break;

    case 'c':
        numProcs = atoi(optarg);
        break;

    default:
        break;
    }
}

I'm tried to figure out what is going on, but I cannot see why argc and argv are not getting values. I have used this exact same parsing code before and it has worked perfectly. Usually segmentation faults are pretty easy (accessing memory locations you shouldn't be), but there is no reason I shouldn't have access to my command line arguments.

Answer 1

You are confused about how to use printf . What you're currently doing is a big no-no:

printf(argv);
printf(argc);

The function expects a const char* string that specifies the format of the output and optional variable argument list. You are passing a double-pointer, which is not a string, and an integer, which is not even a pointer!

You want something like this:

printf( "Number of arguments: %d\n", argc );
for( int i = 0; i < argc; i++ ) {
    printf( "Arg %d: %s\n", i, argv[i] );
}

Read up on printf here: http://en.cppreference.com/w/c/io/fprintf