I want to understand how does array get pass by its value
If I know the length of the array, can I pass its content to the function as I do in my program
Code:#include<stdio.h> void Fun (int content[], int n ); int main () { int array[6] = {6, 5, 4, 3, 2, 1}; Fun (array, 6); } void Fun (int content[], int n ); { }
Yes, but you should still be aware that it is not the array itself that is passed, but rather the array is converted to a pointer to its first element, hence for that function call, the parameter named content is a pointer to the first element of the array named array in the main function.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
The program is not working as it should supposed to do
output : 6422292 6422292 6422292 6422292 6422292 6422292Code:#include<stdio.h> void Fun (int content[], int n ); int main () { int n = 6; int array[6] = {6, 5, 4, 3, 2, 1}; Fun (array, n); } void Fun (int content[], int n ) { for (int i = 0; i < n; i++) { printf(" % d ", content); } }
Try printf(" % d ", content[i]);
> The program is not working as it should supposed to do
If you're not seeing warnings like this, then get a better compiler.Code:$ cat foo.c #include<stdio.h> void Fun (int content[], int n ); int main () { int n = 6; int array[6] = {6, 5, 4, 3, 2, 1}; Fun (array, n); } void Fun (int content[], int n ) { for (int i = 0; i < n; i++) { printf(" % d ", content); } } $ gcc foo.c foo.c: In function ‘Fun’: foo.c:18:11: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=] printf(" % d ", content); ^
If you are seeing warnings like this, then pay attention to them.
If you don't understand a warning, then post the code and the warning.
Don't just run the code anyway and complain that it doesn't work.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Hodor
Try printf(" % d ", content[i]);
It's work
I use MinGW is a port of the free GCC compiler to Windows My compiler is not giving any warning.
There are so many compiler available on internet so will you suggest any good compiler
It should have been given a warning. How are you compiling your programs (e.g. from the command line or using and IDE)?
command gcc -o test test.c
test
command gcc -Wall -Wextra -o test test.c
would be a good place to start.
command gcc --version
would tell us just how old your compiler is.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
gcc --version
gcc (MinGW.org GCC-6.3.0-1) 6.3.0
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
>gcc -Wall -Wextra -o test test.c
warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]
printf(" % d ", content);
Ok, well now you know how to turn warnings onThis might be controversial but when I was at uni if we submitted a program that had a single warning we'd get a zero. Harsh? Not really in my opinion. You should be able to get your program to compile without warnings. So turn the warnings into errors
gcc -Wall -Wextra -Werror -o test test.c
