How is argv a string array when it's a char array?
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How is argv a string array when it's a char array?
Why doesn't argv[0] print the first character of the filename instead of the whole filename string?
argv[0]
If argv is a pointer to an array of chars, then shouldn't accessing it with [n] result with a char? If its a string (as printf(argv[n]) suggests) then why doesn't argv[0][0] get me the first char of filename (compiles but crashes when started)?
argv
[n]
printf(argv[n]
argv[0][0]
int main(int argc, char **argv)
printf(argv[0][0]);
while (1)
return 0;
argv
char
A
char ** can be accessed via two indexing operators like a 2D array (argv[0][0]), but a 2D array is an array of arrays, not an array of pointers. You can find several questions already addressing this here on SO.– John Bollinger
Aug 10 at 12:06
char **
argv[0][0]
@LukaKostic The compiler doesn't care. That's why undefined behavior exists. You make a contract with yourself about that too. The notation
char *argv is simply to indicate to the reader that the intended parameter is indeed an array of pointers, but in the end it's interpreted as char **argv anyway and the compiler really doesn't care.– Iharob Al Asimi
Aug 10 at 12:16
char *argv
char **argv
@LukaKostic There's a good lesson here: code speaks louder than words. Coding is all about the details, which a written out description misses. In this case, you are conceptually correct that
argv[0][0] is the first char of the file name. The problem is that you're not printing that character correctly. Always post code.– John Kugelman
Aug 10 at 12:22
argv[0][0]
@LukaKostic
argv[0][0] by itself doesn't crash. It's the way you used it, that's why the code is so important.– Iharob Al Asimi
Aug 10 at 12:23
argv[0][0]
2 Answers
2
Your code exhibits undefined behavior because you are passing a char to a functions that expects a pointer.
char
To print a single character you have these options,
fputc(argv[0][0], stdout);
putchar(argv[0][0]); // Effectively the same as above
printf("%cn", argv[0][0]);
you can add more of printf() variants.
printf()
The reason your code crashes, is because printf(argv[0][0]); is undefined behavior since the function will try to dereference a pointer but you passed a single character and the value of such character will be interpreted as a memory address.
printf(argv[0][0]);
You really NEED to enable compiler warnings.
How can it get confused about the address? Isnt &argv[0][0] the address and argv[0][0] the char?
– Luka Kostic
Aug 10 at 12:42
@LukaKostic In the end,
char is just an integral type, integers and pointers are interchangeable in c but the behavior of such trade is not always defined. Moreover, if the address was one that is in the range of the memory of your program, it would still be invalid because it's not pointing to a real memory position, or is it? That would not be known beforehand and that's why this is called undefined behavior.– Iharob Al Asimi
Aug 10 at 12:43
char
Clearly, pointers and integers are not interchangable in C, but they are interconvertible. And some compilers perform such conversions even without a cast, which is an extension (though they are typically willing to emit a warning about that).
– John Bollinger
Aug 10 at 13:10
I am using ubuntu server 14 to execute, and gcc compiler to compile the code and argv[n][n] worked for me. Take a look at this code
gcc
argv[n][n]
// C program to illustrate
// command line arguments
#include<stdio.h>
int main(int argc, char* argv)
int counter;
printf("Program Name Is: %s",argv[0]);
if (argc == 1)
printf("nNo Extra Command Line Argument Passed Other Than Program Name");
if (argc >= 2)
printf("nChecking double array elements: %c",argv[0][0]);
return 0;
Output:
Program Name Is: ./test
Checking double array elements: .
What machine are you using to compile and execute? maybe there is some other error... double check!
I needed a printf("%c",argv[0][0]) instead of just printf(argv[0][0])
– Luka Kostic
Aug 10 at 12:38
Yes exactly, if you need to read a
char from a string (array of char) you can do it by %c– Umair
Aug 10 at 13:14
char
char
%c
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Because
argvis a pointer to an array ofcharpointers.– Weather Vane
Aug 10 at 12:01