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For example, the following fragment is incorrect:
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int *p; double f; // p = &f; // ERROR
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This fragment is invalid because you cannot assign a double pointer to an integer pointer That is, &f generates a pointer to a double, but p is a pointer to an int These two types are not compatible (In fact, the compiler would flag an error at this point and not compile your program) Although two pointers must have compatible types in order for one to be assigned to another, you can override this restriction (at your own risk) by using a cast For example, the following fragment is now technically correct:
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int *p ; double f; // p = (int *) &f; // Now technically OK
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The cast to int * causes the double pointer to be converted to an integer pointer However, to use a cast for this purpose is questionable, because the base type of a pointer determines how the compiler treats the data it points to In this case, even though p is actually pointing to a floating-point value, the compiler still "thinks" that p is pointing to an integer (because p is an integer pointer) To better understand why using a cast to assign one type of pointer to another is not usually a good idea, consider the following short program:
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// This program will not work right #include <iostream> using namespace std; int main() { double x, y; int *p; x = 12323; p = (int *) &x; // use cast to assign double * to int * y = *p; // What will this do cout << y; // What will this print return 0; }
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As you can see, p (which is an integer pointer) has been assigned the address of x (which is a double) Thus, when y is assigned the value pointed to by p, y receives only four bytes of data (and not the eight required for a double value), because p is
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an integer pointer Therefore, the cout statement displays not 12323, but a garbage value instead (Try this program and observe the result)
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Assigning Values Through a Pointer
You can use a pointer on the left side of an assignment statement to assign a value to the location pointed to by the pointer Assuming that p is an integer pointer, this assigns the value 101 to the location pointed to by p:
*p = 101;
You can verbalize this assignment like this: "at the location pointed to by p, assign the value 101" To increment or decrement the value at the location pointed to by a pointer, you can use a statement like this:
(*p)++;
The parentheses are necessary because the * operator has lower precedence than the ++ operator The following program demonstrates assignment using a pointer
#include <iostream> using namespace std; int main() { int *p, num; p = # *p = 100; cout << num << ' '; (*p)++; cout << num << ' '; (*p)--; cout << num << '\n'; return 0; }
The output from the program is shown here
100 101 100
Pointer Expressions
Pointers can be used in most valid C++ expressions However, some special rules apply Remember also that you may need to surround some parts of a pointer expression with parentheses in order to ensure that the outcome is what you desire
Pointers
Pointer Arithmetic
There are only four arithmetic operators that can be used on pointers: ++, , +, and To understand what occurs in pointer arithmetic, let p1 be an integer pointer with a current value of 2,000 (that is, it contains the address 2,000) Assuming 32-bit integers, after the expression
p1++;
the contents of p1 will be 2,004, not 2,001! Each time p1 is incremented, it will point to the next integer The same is true of decrements For example,
p1--;
will cause p1 to have the value 1,996, assuming that it previously was 2,000 Here is why: Each time that a pointer is incremented, it will point to the memory location of the next element of its base type Each time it is decremented, it will point to the location of the previous element of its base type In the case of character pointers, an increment or decrement will appear as "normal" arithmetic because characters are one byte long However, every other type of pointer will increase or decrease by the length of its base type You are not limited to only increment and decrement operations You can also add or subtract integers to or from pointers The expression
p1 = p1 + 9;
makes p1 point to the ninth element of p1 s base type, beyond the one to which it is currently pointing While you cannot add pointers, you can subtract one pointer from another (provided they are both of the same base type) The remainder will be the number of elements of the base type that separate the two pointers Other than addition and subtraction of a pointer and an integer, or the subtraction of two pointers, no other arithmetic operations can be performed on pointers For example, you cannot add or subtract float or double values to or from pointers To see the effects of pointer arithmetic, execute the next short program It prints the actual physical addresses to which an integer pointer (i) and a floating-point pointer (f) are pointing Observe how each changes, relative to its base type, each time the loop is repeated (For most 32-bit compilers, i will increase by 4s and f will increase by 8s) Notice that when using a pointer in a cout statement, its address is automatically displayed in the addressing format applicable to the current processor and environment
// Demonstrate pointer arithmetic #include <iostream> using namespace std; int main()
{ int *i, j[10]; double *f, g[10]; int x; i = j; f = g; for(x=0; x<10; x++) cout << i+x << ' ' << f+x << '\n'; return 0; }
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