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We are use to using variables within C without thinking about where they are stored. Most variables are dynamic i.e. can change, therefore they are stored in Ram, unlike a program once developed is static i.e. doesn't change; this is often storage in EPROM or Rom. We can directly access the address where the variable is stored by means of the address operator '&' .This returns the address of the variable followed i.e.
printf("The address of var A is %lx and the contents of var A is %x",&a,a) This address is often fixed by the compiler and cannot be altered easily, very often we want initialize the address of a variable and vary it, C uses the concept of the pointer to handle this. A pointer is a variable which holds an address (This is in fact an address register on the 68Hc11 i.e. X, Y). We can explicitly load up the variable with any number (address which we like) and perform simple mathematical functions on it i.e. add, mul, sub etc. In order to declare the variable as a pointer we need to prefix it with a * in its declarations i.e.
int *point; char *point; float *point;
The type of pointer used describes the size of the data to be read i.e. a char pointer reads 7 bits of data , unsigned char 8 bits i.e. (some programmers use sizeof function to get actual machine size).
char * point == 7 bits unsigned char * point == 8 bits int * point == 15 bits unsigned int * point == 16 bits float * point == 31 bits unsigned float * point == 32 bits
Q. Consider the following page-replacement algorithms. Rank the algorithms on a five-point scale from "bad" to "perfect" according to their page-fault rate. Detach those algorithm
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Note that the parameter for scanf doesn't need the address operators & because name is an address. However the variable name has no defined space. This can cause problems within C
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What is external fragmentation? As process are removed from and loaded to the memory free memory space is bracken into pieces .external fragmentation take place when enough mem
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