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The most appropriate matching for the following pairs

X: m=malloc(5); m= NULL;        1: using dangling pointers
Y: free(n); n->value=5;         2: using uninitialized pointers
Z: char *p; *p = ’a’;           3. lost memory is:

  • A.X—1 Y—3 Z-2
  • B.(X—2 Y—1 Z-3
  • C.X—3 Y—2 Z-1
  • D.X—3 Y—1 Z-2

Answer: D

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