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int n; n = 44; 44 = n; // n is an lvalue / / ERROR: 44 is not an Ivalue // max is an lvalue
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The simplest examples of things that are not lvalues are literals: But symbolic constants are lvalues:
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const int max = 65535; max = 21024;
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even though they cannot appear on the left side of an assignment:
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// ERROR: max is constant
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Lvalues that can appear on the left side of an assignment are called mutabLe Ivalues; those that cannot are called immutable Zvalues. A variable is a mutable lvalue; a constant is an immutable lvalue. Other examples of mutable lvalues include subscripted variables and dereferenced pointers: a[51
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int a[8]; = 22; int* p = &n; *P = 77; // a[5] is a mutable lvalue
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// *p is a mutable lvalue
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Other examples of immutable lvalues include arrays, functions, and references. In general, an lvalue is anything whose address is accessible. Since an address is what a reference variable needs when it is declared, the C++ syntax requirement for such a declaration specifies an lvalue:
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type& refname = lvalue; I
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POINTERS AND REFERENCES
[CHAP. 6
For example, this is a legal declaration of a reference:
int& r = n;
// OK: n is an lvalue
but these are illegal:
int& r = 44; int& r = n++; int& r = cube(n);
// ERROR: 44 is not an hake ERROR: n++ is not an hahe // ERROR: cube(n) is not an lvalue
6.6 RETURNING A REFERENCE A function s return type may be a reference provided that the value returned is an lvalue which is not local to the function. This restriction means that the returned value is actually a reference to an lvalue that exists after the function terminates. Consequently that returned lvalue may be used like any other lvalue; for example, on the left side of an assignment:
EXAMPLE 6.7 Returning a Reference
int& max(int&
m, int& n)
// return type is reference to int // m and n are non-local references
return (m > n m : n);
main0 int m = tout CC max(m,n) tout CC 44, n = 22; " cc max(m,n) CC endl; // changes the value of m from 44 to 55 m cc fl, u cc n CC fl, " CC max(m,n) CC endl;
m cc II, u cc n cc fl,
= 55;
The max ( > function returns a reference to the larger of the two variables passed to it. Since the return value is a reference, the expression max ( m , n > acts like a reference to m (since m is larger than n). So assigning 55 to the expression max (m, n) is equivalent to assigning it to m itself.
EXAMPLE 6.8 Using a Function as an Array Subscript
float& component(float* v, int k)
return v[k-11;
main0 { float v[4]; for (int k = 1; k C= 4; k++) component(v,k) = 1.0/k; for (int i = 0; i < 4; i++) tout CC "v[" CC i CC "1 =
v[i] cc endl;
CHAP. 61
POINTERS AND REFERENCES
The outnut looks like this:
The component ( ) function allows vectors to be accessed using the scientific l-based indexing instead of the default O-based indexing. So assignment component (v, k ) = 1.0 /k is really the assignment v [ k+l ] = 1. 0 /k. We ll see a better way to do this in 9. 6.7 ARRAYS AND POINTERS
Although pointer types are not integer types, some integer arithmetic operators can be applied to pointers. The affect of this arithmetic is to cause the pointer to point to another memory location. The actual change in address depends upon the size of the fundamental type to which the pointer points. Pointers can be incremented and decremented like integers. However, the increase or decrease in the pointer s value is equal to the size of the object to which it points:
EXAMPLE 6.9 Traversing an Array with a Pointer
This example shows how a pointer can be used to traverse an array.
main0 { const int size = 3; short a[size] = (22, 33, 44); tout CC "a = ' CC a CC endl; tout << "sizeof(short) = ' << sizeof(short) << endl; short* end = a + size; // converts size to offset 6 short sum = 0; for (short* p = a; p < end; p++) { sum += *p; tout << '\t p = ' << p; tout << '\t *p = ' << *p; tout C-K "\t sum = " -CC sum -CC endl; > tout << end = I C-K end <=c endl;
The second line of output shows that on this machine short integers occupy 2 bytes. Since p is a pointer to short, each time it is incremented it advances 2 bytes to the next short integer in the array. That way, sum += *p accumulates their sum of the integers. If p were a pointer to doubl e and sizeof(double) were 8 bytes, then each time p is incremented it would advance 8 bytes.
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