Aug 28, 2013 · Recently I have come across this problem which I am unable to understand by myself. What do these three Expressions REALLY mean? *ptr++ *++ptr ++*ptr I have tried Ritchie. But unfortunately was un...

Feb 10, 2016 · In *ptr += 1, we increment the value of the variable our pointer points to. In *ptr++, we increment the pointer after our entire statement (line of code) is done, and return a reference to the variable our pointer points to.

* (ptr++), *ptr++ both result in incrementing the value of pointer but not the +1 expression. Let us say &a [0] i.e address of start of the array = 0x1000 (for simplicity). Let us assume size of each element in array as 4 bytes. x = *ptr++; //x holds 0x1004 : ptr holds 0x1004 x = *(ptr++); //x holds 0x1008 : ptr holds 0x1008 x = *(ptr+1); //x holds 0x100C : ptr still holds 0x1008 (We have not ...

Jul 20, 2013 · If you compile with some warnings on (clang didn't even require any flags), you'll see that your program has three extraneous * operators. Simplifying your crazy-looking expressions yields: ptr++; ++*ptr; ++**ptr; And from that, you should be able to see what's happening pretty clearly: ptr++ just increments ptr, so it points to the second element of p. After this operation ptr - p will always ...

Dec 26, 2014 · As for the question in the title of your post Difference Between *ptr and * (&ptr + 1) then *ptr corresponds to a[0] while using expression *(&ptr + 1) results in undefined behaviour because you are trying to dereference address after ptr.

Feb 28, 2013 · That pointer is then cast to an int*, and that is assigned to ptr. The same could be achieved with int *ptr = &a[5]; in this case. Then ptr - 1 is a pointer pointing sizeof(int) bytes before ptr, that is, to &a[4], and *(ptr - 1) is a[4]. Pointer arithmetic is done in units of "size of pointee".

Feb 19, 2013 · This seems to be the short version of ptr[1], if ptr is an int * pointer of course. so first type cast ptr to int * then increment by one, then dereference.

Jan 24, 2011 · If ptr is a pointer that gives you access to an object at some (arbitrary) position, ptr+1 and ptr-1 will return pointers that give you access to its neighbors.

These statements produce different results because of the way in which the operators bind. In particular, the prefix ++ operator has the same precedence as *, and they associate right-to-left. Thus ++*ptr is parsed as ++(*ptr) meaning "increment the value pointed at by ptr,". On the other hand, the postfix ++ operator has higher precedence than the dereferrence operator *. Thefore *ptr++ means ...

Dec 17, 2012 · int *ptr[10]; is an array of 10 pointers. Reason for segfault: *ptr=a; printf ("%d",*ptr [1]); Here you are assigning the address of array a to ptr which would point to the element a[0]. This is equivalent to: *ptr=&a[0]; However, when you print, you access ptr[1] which is an uninitialized pointer which is undefined behaviour and thus giving ...