解除引用的指针赋值

Question

Consider the code snippet below:

int *p=NULL,*t=NULL,e;
char l=65;
char *k=&l;
p=(int*)&l;
t=(int*)k;
e=*t;
printf("%d\n",*p);
printf("%d\n",*t);
printf("%d\n",e);

The output of the above snippet is as follows:

16705
16705
65

Now it is evident that &l is an address to a char and p=(int*)&l lets an int pointer ie p point to the same location this is the same with pointer t. Now if I dereference p rightly so it will try to dereference 4 consecutive memory blocks (assuming int points to 4 bytes) and print out a garbage value of sorts which happens for *t as well but when I print e I get 65 back which is confusing to me as e=*t which should be same as *p or *t perhaps?

Answer 1

Dereferencing p (ie using *p ) is undefined behavior as there isn't an int at that location. Anything may happen... and no one can explain the output you get.

Compiling your code with "gcc -Wall -Wextra -pedantic ..." gives me:

main.cpp:19:1: warning: array subscript 'int[0]' is partly outside array bounds of 'char[1]' [-Warray-bounds]
   19 | printf("%x\n",*p);

which directly tells you that the program has problems that needs to be fixed.