Command Line arguments in C

[agent_009]

In C, for command line arguments we use

1. int main(int argc, char *argv[])

2. int main(int argc, char *argv[0])

However both of them work fine

What is the exact difference between the two.

Thanx

 

[hr_rezaee]

Hi
as I know
there is no difference.
*argv[]=*argv[0]
regards

 

[sohailshahzad]

both works becase
*argv[]=*argv[0] have the same effect because when nothing is written in [] it means 0.

 

[dragonight_x]

The is no difference between the two line, *argv[]and *argv[0] are exactly the same

 

[sks5440]

When you pass an array you are really passing a pointer to the first element of the array so that *arg[], *arg[1], *arg[999] and **arg are all the same.

 

[rp_istec]

@sks5440,

sorry, but you are compl. wrong.

When you pass an array you are really passing a pointer to the first element of the array so that *arg[], *arg[1], *arg[999] and **arg are all the same.

exmpl.

1. a pointer to the first element of the array, looks like this (datatype *)&arg[0]
2. *arg[1] and esp. **arg ist the same ???, bullschitt !
3. what do you mean, arg[], *arg[1], *arg[999] and **arg are all the same ??, bullschitt !

functionname (int argc, char *argv[]) or functionname (int argc, char *argv[0]) means, argv is a pointer to the first entry of the array with unknown numer of item. Items are of type (char *).
so, the firts entry means, ther is no difference if you write *argv[] or *argv[0]

sorry, Rolf

 

[sandeepsetty27]

Since the argument expects an address and since address of index 0 is same as the base address the value (or address) passed to the main function would be the same. Hence the result.