char concat to string returns wrong length

Question

simple c++ program that adds a char byte to a string. The resulting length is wrong in the output.

#include <iostream>
#include <string>

int main(){

   char x = 0x01;

   std::string test;
   test = x+"test";

   std::cout << "length: " << test.length() << std::endl;
   std::cout << "test: " << test << std::endl;
   return 0;
}

the output:

length: 3
test: est

I am prepending a type byte to the string because I am going to be sending this data through a socket and the other side has a factory that needs to know the type of object to create.

See related question: http://stackoverflow.com/questions/28957950/why-in-the-code-4561-output-is-56 — juanchopanza, May 12 '15 at 14:19

Mohit Jain · Accepted Answer · 2015-05-12T14:16:56.070

4

1 + "test" = "est"  // 1 offset from test

So you are getting the correct answer.

+---+---+---+---+---+
| t | e | s | t | \0|
+---+---+---+---+---+
  +0  +1  +2  +3  +4

What you want is possibly:

std::string test;
test += x;
test += "test";

edited May 12 '15 at 14:16

answered May 12 '15 at 14:15

Mohit Jain

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I want to actually add the char to the string, not move the pointer ahead one.? – user249806 May 12 '15 at 14:16

score 0 · Answer 2 · answered May 12 '15 at 14:15

0

You are not concatenating a char with a std::string as you think you are. This is because "test" is actually a literal const char*, so you are just doing pointer arithmetic when you add x to it. You can replace

test = x + "test";

with

test = std::to_string(x) + "test";

Then your output will be

length: 5
test: 1test

answered May 12 '15 at 14:15

Cory Kramer

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`std::to_string()` is new in C++11. For earlier versions of C++, you can use `std::string(&x, 1)` instead. – Remy Lebeau May 12 '15 at 19:29
@RemyLebeau that will do a different thing – sehe May 12 '15 at 22:00

char concat to string returns wrong length

2 Answers2