C end of line char

I'm having this problem.

char buffer[100];
buffer[0] = "\n";

For some reason, the following statement is true

buffer[0] == 'T开发者_运维问答'

When it should be the "\n" ascii. Why?

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"\n" is a C string, that is a char * pointing at a null terminated series of char elements. Your program takes the address of that string, and stores the lowest 8 bits into buffer[0]. In your case they happen to be the ASCII code for T.

Try the following:

char buffer[100];
buffer[0] = '\n';

'\n' is a char literal, so this will behave as expected.

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Edited: I got it wrong. Your code should be rewritten to the following code:

char buffer[100]={0};
buffer[0]= '\n';

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Try *buffer[0] = '\n'. I think that would give you the desired result, as it is char you are assigning not string. For string use double quotes and for char single quote.

As rightly pointed in the comment buffer[0] is char pointer, so first it needs to be allocated memory too. calloc would be a better choice here as it is going to assign default 0 values whereas malloc will just allocate space having garbage values.