Output of C programs | Set 39 (Pre Increment and Post Increment)

24 Sep, 2021
Prerequisite : Pre-increment and Post increment Question 1  
C
#include <stdio.h> int main(){    char* p = "mayhem";    char c;    int i;    for (i = 0; i < 3; i++) {        c = *p++;    }    printf("%c", c);     return 0;}
yhea
y
h
e
a
Answer : y 
Explanation: The pointer ‘p’ points at the third location of the character array. The reason is that in the ‘for’ loop iteration, the value of the character pointer ‘p’ has been incremented thrice.Question 2  
CPP
#include<stdio.h>void test(char c[]){    c=c+2;    c--;    printf("%c",*c);}int main(){    char ch[5]={'p','o','u','r'};    test(ch);         return 0;}
What is the output of the above program?  
pour
p
o
u
r
 
Answer: o 
Explanation: When the character array ‘ch’ is passed as an argument to the function ‘test()’, the base address of the array ‘ch[]’ is passed. The variable ‘c’ in the function ‘test()’ points at the second location of the array. And then it decrements by 1 pointing to ‘o’.Question 3  
C
#include<stdio.h>int main(){    int i;    char ch[] = {'x', 'y', 'z'};    char *ptr, *str1;    ptr = ch;    str1 = ch;    i = (*ptr-- + ++*str1) - 10;    printf("%d", i);         return 0;}
What is the output of the above program if the ASCII values of characters ‘x’=120, ‘y’=121, ‘z’=122?  
231233232363
231
233
232
363
 
Answer : 231
Explanation: The pointer ptr points to ‘x’. Step1: Since, it is a post-decrement operation, hence the value remains 120 and is decremented later. Step2 :The pointer str1 points at ‘x’. Since ++ and * have same precedence level evaluation starts from Right to Left. For the expression ++*str1, Firstly *str1 is evaluated which gives 120 i.e. ASCII Equivalent of x. Next we evaluate the prefix increment ++120 = 121. Hence the final result is (120+121)-10=131
Question 4 – What will be the output of following Program?  
CPP
#include<stdio.h> int main(void){    char *ptr = "Linux";    printf("\n [%c] \n", *ptr++);    printf("\n [%c] \n", *ptr);     return 0;}
Output :  
    [L] 

    [i]
Explanation : Since the priority of both ‘++’ and ‘*’ are same so processing of ‘*ptr++’ takes place from right to left. Going by this logic, ptr++ is evaluated first and then *ptr. So both these operations result in ‘L’. Now since a postfix ‘++’ was applied on ptr so the next printf() would print ‘i’.Question 5 – What will be the output of following Program?  
CPP
#include <stdio.h>int main(){        int num1 = 5;        int num2 = 3;        int num3 = 2;        num1 = num2++;        num2 = --num3;        printf("%d %d %d", num1, num2, num3);        return 0;}
231311327321
231
311
327
321
Answer:311 
This article is contributed by Brahmani Sai. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
DhineshJ
chenosebu
C-Output
Program Output
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Output of C++ programs | Set 44
Runtime Errors
Different ways to copy a string in C/C++
Output of C++ Program | Set 1
error: call of overloaded ‘function(x)’ is ambiguous | Ambiguity in Function overloading in C++
Output of Java Programs | Set 12
Output of Java program | Set 5
Output of Java Program | Set 6
Output of Java program | Set 15 (Inner Classes)
Output of Java Programs | Set 48 (Static keyword)