CODEWITHSUNDEEP
ccommand-line-argumentsargv
nonenone
nonenone

Reputation: 130

result of passing argv variable to main in this format main( int argc, char const * argv )

I am trying to figure out what will be result of passing argv variable to main in this format. main( int argc, char const * argv ).

I know right way of using it is main( int argc, char const **argv) or main( int argc, char const *argv[]).

Here is my code snippet, 

#include <stdio.h>

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

Here is output,

$ ./a.exe

..

$ ./a.exe 1

2

....

$ ./a.exe 1 2

2

...

Does it fetch from whatever, argv points? Why does it terminate before even reaching argc.

Upvotes: 0

Views: 740

Answers (1)

Sourav Ghosh
Sourav Ghosh

Reputation: 134326

It causes undefined behavior as the supplied type and the expected type does not match thereby this is a clear case of constraint violation.

Quoting C11, chapter chapter §5.1.2.2.1/p2, (emphasis mine)

If they are declared, the parameters to the main function shall obey the following constraints:

  • ...

  • If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings, which are given implementation-defined values by the host environment prior to program startup

You can re-write char * argv[] as char **argv NOTE considering the array decay property, but a char *[] and a char * are not compatible in any way.

NOTE:

Quoting C11, chapter §5.1.2.2.1, footnote 10)

Thus, int can be replaced by a typedef name defined as int, or the type of argv can be written as char ** argv, and so on.

Upvotes: 2

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