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I've just started to learn C programming. When coming to the String, I get confused with the function 'strcpy'. I tried switching places of first argument and the second argument. When I run the program, it just shows a 'S'. What does that mean?

char s2[ ]= "Hello";
char s1[10];
strcpy(s2, s1);
printf("Source string = %s\n", s2);
printf("Target string = %s\n", s1);

I thought the output would be null. But it just shows a 'S'.

Uwe Keim
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Vincent.N
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2 Answers2

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Based on the printf statements, you have the arguments to strcpy mixed up.

As it is now, you're copying s1 to s2. The array s1 is uninitialized however, so the values it contains are indeterminate.

To copy s2 to s1, switch the parameters:

strcpy(s1, s2);

If you leave it as is, you need to explicitly set s1 to an empty string to get consistent results.

char s1[10] = "";
dbush
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  • Was the 'S' letter which is the output of the my code above just the random thing or it has any special meanings? Thank you for the answer. – Vincent.N Feb 03 '19 at 16:13
  • @Vincent.N Since `s1` was not initialized, you read whatever random values happened to be there. – dbush Feb 03 '19 at 16:17
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In C, strings are zero terminated. This means that an empty string is a string containing a single "zero terminator" char.

When an empty string is copied, the single "zero terminator" char is copied. The destination string still has an address (the "pointer to the strings chars" will point to the zero terminator) and the pointer to the string will not be NULL.

Brendan
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