Execution of printf() and Segmentation Fault

Question

#include

int main()
{
    char *name = \"Vikram\";
    printf(\"%s\",name);
    name[1]=\'s\';
    printf(\"%s\",name);
    return 0;
}

Answer 1

The reason you are getting a segmentation fault is that C string literals are read only according to the C standard, and you are attempting to write 's' over the second element of the literal array "Vikram".

The reason you are getting no output is because your program is buffering its output and crashes before it has a chance to flush its buffer. The purpose of the stdio library, in addition to providing friendly formatting functions like printf(3), is to reduce the overhead of i/o operations by buffering data in in-memory buffers and only flushing output when necessary, and only performing input occasionally instead of constantly. Actual input and output will not, in the general case, occur at the moment when you call the stdio function, but only when the output buffer is full (or the input buffer is empty).

Things are slightly different if a FILE object has been set so it flushes constantly (like stderr), but in general, that's the gist.

If you're debugging, it is best to fprintf to stderr to assure that your debug printouts will get flushed before a crash.