why segmentation fault in second code?

Question

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Why is this C code causing a segmentation fault?

char array[6] = "kapil";  
array [0] =  'K';    

this code runs ok.
but following code produce segmentation fault ?

char* array = "kapil";  
array [0] =  'K';   

why first does not produce seg-fault.

Answer 1

char* array = "kapil"; declares a string literal and is equivalent to const char*. This may well exist in read-only memory and cannot be modified.

Strictly speaking, attempts to modify it result in undefined behaviour but in practice a seg fault is likely.

Answer 2

char* array = "kapil"; using this type of declaration, it copies kapil in read-only context, you can change its contents.