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mainResult is set to the address of the automatic Matrix object res Unfortunately, the storage of res is deallocated on completion of trouble() On the return to main(), mainResult is addressing technically unallocated memory (In this example, the address may still remain valid because we have not yet invoked another function to overlay some or all of the activation record allocated to support trouble(), making such an error more difficult to detect) Using mainResult later in main() will yield surprising results However, passing the address of main()'s automatic object m1 to the function trouble() is always safe The storage for main() is guaranteed to remain on the stack for the duration of the call to trouble(), and the memory for m1 remains accessible for the duration of the call to trouble() When the address of an automatic object is stored in a pointer with a lifetime longer than that of the automatic object, the pointer is said to be a dangling pointer This is a serious programmer error, because the contents of the addressed memory are unpredictable If the bits at that address are somehow relevant (so that the program does not generate a segment fault, for example), the program may run to completion but provide invalid results Register Automatic Objects Automatic objects used heavily within a function can be declared with the keyword register If possible, the compiler will load the object into a machine register If it cannot, the object remains in memory Array indexes and pointers occurring within a loop are good candidates for register objects

for ( register int ix = 0; ix < sz; ++ix ) // for (register int *p = array ; p < arraySize; ++p ) //

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A parameter can also be declared as a register variable:

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bool find( register int *pm, int val ) { while ( *pm ) if ( *pm++ == val ) return true; return false; }

Register variables may increase the speed of a function if the variables selected are used heavily Using the keyword register is only a hint to the compiler Some compilers may ignore this hint and use register allocation algorithms to figure out the best candidates to be placed within the available machine registers Because the compiler is aware of the machine architecture on which the program is run, it is often able to make a more informed decision when selecting the content of machine registers Static Local Objects It is also possible to declare, within a function definition or within a compound statement of a function definition, a local object that will persist for the entire duration of the program When the value of a local object must persist across invocations of the function, an ordinary automatic object cannot be used The value of an automatic object is discarded each time the function exits The solution in this case is to declare the local object as static A static local object has static storage duration, or static extent Though its value persists across invocations of its function, the visibility of its name remains limited to its local scope A static local object is initialized the first time the execution of the program passes through the object's declaration For example, here is a version of gcd() that traces the depth of its recursion using a static local object:

R5 = (16, 13) LUT (1/23) mod m i 27, 24 R* = (29, 22) 5 RNS subtractor RNS multiplier (22, 22) x4 = 22

#include <iostream> int traceGcd( int v1, int v2 ) { static int depth = 1; cout "depth #" depth++ endl; if ( v2 == 0 ) { depth = 1; return v1; } return traceGcd( v2, v1%v2 ); }

The value associated with the static local object depth persists across invocations of traceGcd() The initialization is performed only once: when traceGcd() is first called The following small program exercises traceGcd():

#include <iostream> extern int traceGcd(int, int); int main() { int rslt = traceGcd( 15, 123 ); cout "gcd of (15,123): " rslt endl; return 0;

0.01 0.1

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When compiled and executed, the program generates the following results:

depth #1 depth #2 depth #3 depth #4 gcd of (15,123): 3

An uninitialized static local object is automatically initialized to 0 by the program In contrast, automatic objects have arbitrary values unless they are explicitly initialized The following program illustrates the difference between the default initialization of automatic and static local objects and the danger of not initializing automatic objects

0.0001 0.001

#include <iostream> const int iterations = 2; void func() { int value1, value2; // uninitialized static int depth; // implicitly initialized to 0 if ( depth < iterations ) { ++depth; func(); } else depth = 0; cout "\nvalue1:\t" value1; cout "\tvalue2:\t" value2; cout "\tsum:\t" value1 + value2; } int main() { for ( int ix = 0; ix < iterations; ++ix ) func(); return 0; }

The output of its execution is as follows:

RNS to mixed-radix conversion (Example 14.1).

value1: value1: value1: value1: value1: value1:

0 0 0 148620 2147479844 0

0.01 0.01

value2: value2: value2: value2: value2: value2:

74924 68748 68756 2350 671088640 68756

sum: sum: sum: sum: sum: sum:

74924 68748 68756 150970 1476398812 68756

0.1 0.1 0.1

Notice that value1 and value2 are uninitialized automatic objects Their initial values, as shown in the output here, are completely random, and the result of their sum is unpredictable However, even though depth is uninitialized, its initial value is guaranteed to be 0, ensuring that func() recursively invokes itself only twice

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.

R1(s..1) R1(1)