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Editorial Team
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Asked: 2026-06-04T10:49:31+00:00

char str[]=Hello; this allocates 6 bytes for the string , but if i write

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char str[]="Hello";

this allocates 6 bytes for the string , but if i write

char *str = "Hello";

will this overwrite data because it was just meant to store 1 char? So what i’m asking is that when i declare a string, but not initialize it (char str[12]; ) , do 12 bytes get reserved here or when i initialize it? And if they do get initialized here, so that means that in:

char *str;

only 1 byte gets reserved, but when i make it point to a string, doesn’t that overwrite data beyond it’s bounds?

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1 Answer

  1. Editorial Team
    char *str;

    does not reserve any data for content. It is a pointer, sized to hold a memory address.

    char *str = "Hello";

    6 bytes for { 'H', 'e', 'l', 'l', 'o', 0 } already has been stored somewhere by the compiler. Now you are making a variable holding its address (pointing to it). The string content is not copied.

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