If you invoke the program as follows:
program.exe 1.dat 2.dat
Then the main() routine of program.exe will receive argc, and argv as follows:
argc = 3
argv[0] = "program.exe"
argv[1] = "1.dat"
argv[2] = "2.dat"
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Question
file copying in C
hello I am looking how to copy a file in C
I mean
like this :
in ms dos prompt I will prompt
program.exe 1.dat 2.dat
and 1.dat will be copied to 2.dat ( I dont know how to use argc argv , I would appreciate if you can explain this)
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Answers
#include <stdio.h>
#define buflen 1000
char buffer[buflen];
int main( int argc, char *argv[] ){
FILE *F,*T;
int n;
if( argc != 3 ){
printf("usage: %s from to\n",argv[0]);
exit(1);
}
if( !(F=fopen(argv[1],"r")) ){
perror(argv[1]);
exit(1);
}
if( !(T=fopen(argv[2],"w")) ){
perror(argv[2]);
exit(1);
}
while( n = fread(buffer,1,buflen,F) ){
fwrite(buffer,1,n,T);
}
}
But the stdio library takes care of matching buffer sizes to file systems' preferred block sizes, so the buffer size you use in a stdio-using program makes no difference in the size and alignment of the read and write transactions with the actual disk device.
But I, too, would have chosen a larger buffer size.
And, if you're looking for efficiency, I'd probably forego the use of stdio calls entirely.
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by: theMuzzPosted on 2003-10-22 at 12:49:32ID: 9601636
/* argc will give you the number of arguments you've passed
argv[] returns the values for each argument
int main(int argc, char* argv[])
{
int numberOfArguments;
numberOfArguments = argc;
string arg1, arg2, arg3; // and so on
arg1 = argv[0];
arg2 = argv[1];
arg3 = argv[2];
}
Can't remember if first argument is argv[0] or argv[1], I'm pretty sure it's 0.
so... program.exe three small arguments
would result in
arg1 = 'three'
arg2 = 'small'
arg3 = 'arguments'
Hope this helps