What is the output of the following code snippet?int x = 5;int y = 10;int *ptr = &x;*ptr = y;printf("%d", x);510015Generates a compilation error
Question
What is the output of the following code snippet?int x = 5;int y = 10;int *ptr = &x;*ptr = y;printf("%d", x);510015Generates a compilation error
Solution
The output of the code snippet will be 10.
Here's the step by step explanation:
-
int x = 5;This line declares an integer variablexand assigns it the value 5. -
int y = 10;This line declares another integer variableyand assigns it the value 10. -
int *ptr = &x;This line declares a pointer variableptrand assigns it the address ofx. So,ptris now pointing tox. -
*ptr = y;This line assigns the value ofyto the locationptris pointing to. Sinceptris pointing tox, this effectively changes the value ofxto 10. -
printf("%d", x);This line prints the value ofx, which is now 10. So, the output of this code snippet is 10.
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