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Home/ Questions/Q 7003487
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T21:04:30+00:00 2026-05-27T21:04:30+00:00

This seems like it should be easy but i’ve spent way too much time

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This seems like it should be easy but i’ve spent way too much time on it. Hopefully someone can help.

char *labels[] = { "apple", "orange" }; // each items inside label is string literal. We can't change them-

Look at the below

char a[]="hi";
char b[]="hello";
char *name[]={a,b};//each item inside name is not string literal rite??
*name="bye";
puts(a);

I thought output will be bye since i changed the content of a[] using *name="bye"

But the output is still hi. why?

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  1. Editorial Team
    Editorial Team
    2026-05-27T21:04:31+00:00Added an answer on May 27, 2026 at 9:04 pm

    char *name[]={a,b};//each item inside name is not string literal rite?? – not quite, each element is a pointer to a char.

    So name is an array of pointers. *name="bye" changes what the first of those pointers change to. It doesn’t change the memory that the old pointer pointed to.

    If you wanted to do that you would use strncpy in general, but note that here you don’t have a large enough array to do that.

    (I would also have expected your compiler to give you a warning about assigning a const char * to a char* for *name="bye")

    Before *name="bye":

    State before

    After *name="bye":

    State after

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