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Does the plain literal refer to the address and *literal refer to the actual value at the address? So, AFTER:
int i = 0;
int *iPointer = &i;
the following expression will lookup the VALUE AT memory address &i:
*iPointer
and the following will simply yield the memory address &i:
iPointer
I stepped through and verified my hypothesis, but I want to make sure (you never know with these things).
I guess I'm just confused by the * symbol's different purposes in declaration and access.
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When declaring a pointer type, place the asterisk next to the type name. Although you generally should not declare multiple variables on a single line, if you do, the asterisk has to be included with each variable.
Here, * is called dereference operator. This defines a pointer; a variable which stores the address of another variable is called a pointer. Pointers are said to point to the variable whose address they store.
Pointers must be declared before they can be used, just like a normal variable. The syntax of declaring a pointer is to place a * in front of the name. A pointer is associated with a type (such as int and double ) too.
Simply remembering to put one asterisk per pointer is enough for most pointer users interested in declaring multiple pointers per statement.
Yes, that's absolutely correct.
Also, note that & can also declare a reference, depending on context.
In a declaration, * declares a pointer type.
When applied on a pointer (like *ptr), it represents the dereference operator and returns the value the pointer points to.
Note that operator * can be overloaded, so you can also apply it to objects, not only pointers, and have it do whatever you want.
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