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Harakim · Jun 5th, 2006, 6:52 PM

Although this isn't technically true "char *name" and "char name[xx]" are basically the same.

When you are passing an array in C, you can pass it as a pointer and dereference it with brackets.

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int foo( char *bar )
{
	bar[1] = 100;
}


int foo2()
{
	char bar2[20];
	
	foo( bar2 );	

	//bar2[1] now equals 100
}

I have heard that they are not the same at all, so hopefully someone will be enraged by my saying they are and come and explain it. As for passing arrays though, I'm pretty sure that should work.

(Toggle Plain Text)

int foo( char **bar )
{
	bar[1][2] = 100;
}


int foo2()
{
	char bar2[20][10];
	
	foo( bar2 );	

	//bar2[1][2] now equals 100
}

EDIT: fixed error in code

Jun 5th, 2006, 6:52 PM	#21
Harakim Hobbyist Programmer Join Date: May 2006 Location: West Jordan, Utah, United States Posts: 176 Rep Power: 3	Although this isn't technically true "char name" and "char name[xx]" are basically the same. When you are passing an array in C, you can pass it as a pointer and dereference it with brackets. (Toggle Plain Text) int foo( char bar ) { bar[1] = 100; } int foo2() { char bar2[20]; foo( bar2 ); //bar2[1] now equals 100 } int foo( char bar ) { bar[1] = 100; } int foo2() { char bar2[20]; foo( bar2 ); //bar2[1] now equals 100 } I have heard that they are not the same at all, so hopefully someone will be enraged by my saying they are and come and explain it. As for passing arrays though, I'm pretty sure that should work. (Toggle Plain Text) int foo( char bar ) { bar[1][2] = 100; } int foo2() { char bar2[20][10]; foo( bar2 ); //bar2[1][2] now equals 100 } int foo( char *bar ) { bar[1][2] = 100; } int foo2() { char bar2[20][10]; foo( bar2 ); //bar2[1][2] now equals 100 } EDIT: fixed error in code