Why does these two code pieces give different results

Question

I was trying to learn about "<" operator on c++ strings and tried some test cases. I realized the two codes which I thought should behave the same was giving different results. Below are the codes, what is the reason for this?

  string s="test";
  string k="tes";
  cout<<(s<k)<<endl; //returns false so it prints 0
  cout<<("test"<"tes")<<endl; // returns true so it prints 1

score 4 · Accepted Answer · answered May 08 '20 at 21:02

4

(s < k) compares the values of the strings as you would expect.

("test" < "tes") compares the pointers to the beginning of the string literals as the compiler decides to arrange them in memory. Therefore, this comparison may return 0 or 1 depending on the compiler and settings in use, and both results are correct. The comparison is effectively meaningless.

The "C way" to compare these string literals would be strcmp("test", "tes").

answered May 08 '20 at 21:02

cdhowie

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Then does the string literal "test" does not really refer to a string object? Is it like type casting when I declare string s ="test"; ? – OgiciBumKacar May 08 '20 at 21:03
1

@OgiciBumKacar Correct, that invokes the `std::string(char const *)` implicit conversion constructor. – cdhowie May 08 '20 at 21:04
Thank you! This clears a lot – OgiciBumKacar May 08 '20 at 21:04

score 0 · Answer 2 · answered May 08 '20 at 21:05

s and k are string objects for which a comparison operator has been defined and performs what you expect.

"test" and "tes" are pointers to char that hold the address of the locations where these characters are stored. Thus the comparison is on the addresses.

Why does these two code pieces give different results

2 Answers2