Difference between Array initializations

Question

Please see the following statements:

char a[5]="jgkl"; // let's call this Statement A
char *b="jhdfjnfnsfnnkjdf"; // let's call this Statement B , and yes i know this is not an Array
char c[5]={'j','g','k','l','\0'}; // let's call this Statement C

Now, is there any difference between Statements A and C? I mean both should be on Stack dont they? Only b will be at Static location.

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So wouldn't that make "jgkl" exist at the static location for the entire life of the program? Since it is supposed to be read-only/constant? Please clarify.

Answer 1

No, because the characters "jgkl" from Statement A are used to initialize a, it does not create storage in the executable for a character string (other than the storage you created by declaring a). This declaration creates an array of characters in read-write memory which contain the bytes {'j','g','k','l','\0'}, but the string which was used to initialize it is otherwise not present in the executable result.

In Statement B, the string literal's address is used as an initializer. The variable char *b is a pointer stored in read-write memory. It points to the character string "jhdfjnfnsfnnkjdf". This string is present in your executable image in a segment often called ".sdata", meaning "static data." The string is usually stored in read-only memory, as allowed by the C standard.

That is one key difference between declaring an array of characters and a string constant: Even if you have a pointer to the string constant, you should not modify the contents.

Attempting to modify the string constant is "undefined behavior" according to ANSI C standard section 6.5.7 on initialization.

Answer 2

If a[] is static then so is c[] - the two are equivalent, and neither is a string literal. The two could equally well be declared so that they were on the stack - it depends where and how they are declared, not the syntax used to specify their contents.