![]() Dereferencing the expression *( *( a + i ) + j ) you will get the j-th element of the i-th row of the array. The expression *( a + i ) + j points to the j-th element of the "row" of the two-dimensional array. The expression *( a + i ) has the type int that at once is implicitly converted to a pointer of the type int * to its firs element in the enclosing expression. So the expression a + i has the type int ( * ). Dereferencing the expression you will get the i-th row (one-dimensional array) that in turn used in expressions is converted to a pointer to its first element. If you want that a pointer point to the i-th element of the array you need to write the expression a + i. So a pointer to such object has the type int ( * ). Then the array designator a is converted to a pointer to its first element ("row"). So if you have an array declared like: int a Structured Level Language: C is a structured programming language that allows a complex program to be divided into simpler programs called functions. Here is a demonstration program: #include Īrray designators used in expressions (with rare exceptions) are implicitly converted to pointers to their first elements. What are the advantages of C Some of the advantages of C are: Middle-Level Language: As C is in the middle of a high-level language and low-level language, it has features of both. The expression x is evaluated like: *( *( x + y ) + z ) So, for a multidimension array, x as a 2 dimension array, x Pointer arithmetic of the int* + j gives the address of the integer, which is then finally de-referenced to give the actual int.We get a pointer to the first element, type int*. Again, there is array decay on this one, because it too is an array part of an expression.Pointer arithmetic on that array pointer type yields array number i of type int.So we get *((arr)+(i)), where arr is decayed into a pointer to the type of the first element, int(*).In the expression arr, the is just "syntactic sugar" for pointer arithmetic (see Do pointers support "array style indexing"?).Why typing out the & as done in your pseudo code is just confusing.Ī practical example would be: int arr Except for a few exceptions, namely the & address of and sizeof operators. so when i was calling answer0count (which is a global variable), thats where i was getting myself into trouble. When part of an expression, arrays always decay into a pointer to it's first element. Re: help - compiling error: invalid indirection array problems well, i figured it out - i had a variable in that same function called answer. This array too "decays" into a pointer of the first element, so you get: ( (*((x)+(y))) + (z) ) So *((x)+(y)) gives you that array, x decays into a pointer to the first element, which is then de-referenced to give you array number y. So, for a multidimension array, x as a 2 dimension array, x would be equivalent to (((&(x))+(y))+(z))
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