Select the correct answerPredict the output produced by below given code snippet in C.#include<stdio.h>void Function(char *S1, char *S2){ char *Buffer = S1; S1 = S2; S2 = Buffer;} int main(){ char *S1 = "Code Tantra"; char *S2 = "Coding Castle"; Function(S1, S2); printf("%s,%s", S1, S2); return 0;}Optionschar *S1 = "Code Tantra"; char *S2 = "Coding Castle"; these statements throw segmentation faultThis code prints "Code Tantra,Coding Castle"This code prints "Code Tantra,Code Tantra"This code prints "Coding Castle,Code Tantra"Code throws compile time error
Question
Select the correct answerPredict the output produced by below given code snippet in C.#include<stdio.h>void Function(char *S1, char *S2){ char *Buffer = S1; S1 = S2; S2 = Buffer;} int main(){ char *S1 = "Code Tantra"; char *S2 = "Coding Castle"; Function(S1, S2); printf("%s,%s", S1, S2); return 0;}Optionschar *S1 = "Code Tantra"; char *S2 = "Coding Castle"; these statements throw segmentation faultThis code prints "Code Tantra,Coding Castle"This code prints "Code Tantra,Code Tantra"This code prints "Coding Castle,Code Tantra"Code throws compile time error
Solution
The correct answer is "This code prints "Code Tantra,Coding Castle"".
Here's the step by step explanation:
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The main function initializes two pointers, S1 and S2, to point to the strings "Code Tantra" and "Coding Castle" respectively.
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The Function is called with S1 and S2 as arguments. Inside the Function, a new pointer Buffer is created and set to point to the same string as S1.
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Then, S1 is set to point to the same string as S2, and S2 is set to point to the same string as Buffer (which is the original string of S1). However, these changes are local to the Function, because S1 and S2 are passed by value, not by reference. This means that the original pointers in the main function are not affected by these changes.
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After the Function is finished, the printf statement in the main function prints the strings that S1 and S2 point to, which are still "Code Tantra" and "Coding Castle".
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