C: How to open a file using the parameter to main argv?

Question

I've been trying to open a file using the char** argv parameter. But unfortunately, I stumped upon a problem reading the path to the file when I pass it in this format: program.exe Function SourceFile DestFile.

I'm using notepad++ to write the code and GCC to compile and pass the arguments to the function

UPDATE: I fixed the code, should be working right now...

#include <stdio.h>
#include <string.h>

void textCopy(FILE* sourceFile, FILE* destinationFile);
void binaryCopy(FILE* sourceFile, FILE* destinationFile);

int main(int argc, char** argv)
{
    printf(argv[2]);
    if ((strcmp(argv[1], "textCopy") != 0 && strcmp(argv[1], "binaryCopy") != 0))
    {
        printf("Error: Function Doesn't Exist");
        return 1;
    }
    FILE* sourceFile = fopen(argv[2], "r");
    if (sourceFile == NULL)
    {
        printf("Error: Source File Doesn't Exist");
        return 1;
    }
    FILE* destinationFile = fopen(argv[3], "w");
    if (destinationFile == NULL)
    {
        printf("Error: Destination File Doesn't Exist");
    }
    if (strcmp(argv[1], "textCopy") == 0)
    {
        textCopy(sourceFile, destinationFile);
    }
    else
    {
        binaryCopy(sourceFile, destinationFile);
    }
    getchar();
    return 0;
}

void textCopy(FILE* sourceFile, FILE* destinationFile)
{
    char letter = 0;
    while (letter != EOF)
    {
        letter = fgetc(sourceFile);
        fputc(letter, destinationFile);
    }
}

void binaryCopy(FILE* sourceFile, FILE* destinationFile)
{
    printf("Ignore");
}

I searched for solutions in the internet but nothing seemed to work out, when i read argv[2] i only get the C:\ part from the path and not the whole path...

Thanks!

Answer 1

Try something in the spirit of:

program.exe textCopy "C:\path to\the source file\my source file.ext" "C:\path to\the destination file\my destination file.ext"

It's the double quotes around the file path in the command line you're missing.