Explaining (int argc, char * argv[])

You all have often came across this term while going through a c or c++ code.I just want to explain this in brief especially for the newbies who find it difficult to understand this concept.

Here argc represents the number of arguments(command line argument) and argv is the argument vector pointing it.
Suppose I am compiling a c code ‘test.c’ so what i give at command line is

$ gcc test.c -o test
$ ./test                                                           //for running executable file test

so here argc=1 as no of command line arguments here is one and argv[0]=./test

similarly if executable is run as

$ ./test <arg1> <arg2>

then here argc=3 and argv[0]=./test ,argv[1]=arg1 and argv[2]=arg2

Below is the c code i have written try it for better understanding of this concept.Have lots of fun with c codes 🙂

#include
main(int argc, char* argv[])
{
int i;
printf("argc = %d\n", argc);
for (i = 0; i < argc; i++)
printf("argv[%d] = %s\n", i, argv[i]);
}
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