Examine the following code. what will be the output from the “printf()” statement below?
#define s_size 10
char s[10] = { ‘g’ , ’o’ , ’m’ , ’i’ , ’z’ , ’z’ , ’o’ , ’u’ };
int j = 0;
for (j = s_size -2; j > 4; { printf( “%c”, s[j]); }
using the definitions below (ignore all compile errors):
char s1 [ 50 ] = “jack”, s2 [ 50 ]=”fred”;

uoz (with one space on the left most side)

Explanation:

#define is used to define macros, it just substitutes the value in the program.

For example:

#define S_SIZE 10

so, it put the value 10 where we write 'S_SIZE'.

then, character array define which has 8 elements.

Note: Array index starts from 0.

the, for loop is execute from 8 to j > 4( means 5).

it prints s[8] which not present in the array so, it prints blank space.

s[7] it prints 'u'

s[6] it prints 'o'

s[5] it prints 'z'

then the loop will be terminated.

Therefore, the answer is uoz with space on the leftmost side.

answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation: