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Home/ Questions/Q 8580965
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T20:56:18+00:00 2026-06-11T20:56:18+00:00

How does this work:: char Test1[8] = {abcde} ; AFAIK, this should be stored

  • 0

How does this work::

char Test1[8] = {"abcde"} ;

AFAIK, this should be stored in memory at Test1 as

a b c d e 0 SomeJunkValue SomeJunkValue

instead it get stored as:

a b c d e 0 0 0

Initializing only adds one trailing NULL char after the string literals but how and why all other array members are initialized to NULL ?
Also, any links or any conceptual idea on what is the underlying method or function that does:char TEST1[8] = {"abcde"} ; would be very helpful.
How is:

char Test1[8] = {"abcde"} ;

different from

char Test1[8] = "abcde" ;

?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T20:56:19+00:00Added an answer on June 11, 2026 at 8:56 pm

    Unspecified members of a partially initialized aggregate are initialized to the zero of that type.

    6.7.9 Initialization

    21 – If there are fewer initializers in a brace-enclosed list than there are elements or members
    of an aggregate, or fewer characters in a string literal used to initialize an array of known
    size than there are elements in the array, the remainder of the aggregate shall be
    initialized implicitly the same as objects that have static storage duration.

    10 – […] If an object that has static or thread storage duration is not initialized
    explicitly, then:

    • if it has pointer type, it is initialized to a null pointer;
    • if it has arithmetic type, it is initialized to (positive or unsigned) zero; […]

    For the array char Test1[8], the initializers {"abcde"} and "abcde" are completely equivalent per 6.7.9:14:

    An array of character type may be initialized by a character string literal or UTF−8 string literal, optionally enclosed in braces.

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