Self Made Doubt

Question

if we write in C

char a[]=”TAJ”;

char b[]=”TAJ”;

and on comparing (a==b) false is returned

where as in char *p=”TAJ”;

char *q=”TAJ” it return TRUE

why??

Comments

what should printing a and b return? without de-referencing or anything, what value do they exactly hold?

 

Second part :
“according to the standard, they are not required to have the same address, it's unspecified. But in practice, most compiler will make identical string literals holding the same address”. https://stackoverflow.com/questions/19442533/read-only-string-in-c

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@Sagar475 See this,

char a[] = “Taj”. here a represent array that has a base address say 100.

and b[] = “Taj” here b represent array with base address say 200

so if(a==b) is false

but,

char *p = “Taj”, here p represent pointer which store the base address of character array “Taj” says the base address is 500.

similarly char *q = “Taj”, here q is also char pointer which stores the base address of char array “Taj” so store 500.

so if (p==q) returns true.

Note: char *p = “TAj” when writing we create a character array in the data section which allocate at compile time. So both p and q store same address.

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@Yaman Sahu “Taj” – is a string literal. Is it always guaranteed that two string literals (having same characters) compare equal in C?

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@Arjun sir
yes sir.

This code I am executing in my PC 

compiler: gcc 64 bit

#include<stdio.h>

int main()
{

    char *p = "Taj";
    char *q = "Taj";
    printf("p = %u, q = %u", p, q);
    return 0;
}

// o/p : p = 4210736, q = 4210736

 

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Thats just one output of one compiler in one system :) 

You can see the below rule in this section of C standard.

• Whether two string literals result in distinct arrays (6.4.5).

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Yes sir, I realized my mistake.

Thank you, sir :)

Answer 1

In the First Case, you are comparing the address of the two array which will always be different. Here value of a = 1000 (address 1), b = 2000 (address 2)  and as we know they cannot be stored on the same address therefore, the result is false.

In the Second Case, you are making a string in a memory and you are pointing to that memory, like TAJ is stored in consecutive memory location suppose starting from address 1000, i.e. T → 1000 ( address ). this type of declaring is also called “string constant”.

So, in the Second Case, as we are using the same string, a & b both point to the same string present in the memory (compiler optimization). Therefore there pointing address is the same, making the result is true.

 NOTE: the variables a and b are present at different memory location, they just point to the same memory location.