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POunspecified ouput using inline function in c++
 

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  1. POunspecified ouput using inline function in c++
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    <p>I was playing with code to understand internal and external linkage in c++.I came up with the code whose out seems to vary depending on the sequence in which it is linked.</p> <p><br><strong>test1.cpp</strong> <br/></p> <pre><code>#include&lt;iostream&gt; using namespace std; inline int c() { static int p=0; p++; return p; } void a() { cout&lt;&lt;"\nIn function a() , c = "&lt;&lt;c(); } </code></pre> <p><br><strong>test2.cpp</strong><br/></p> <pre><code>#include&lt;iostream&gt; using namespace std; inline int c() { static int p=12; p++; return p; } void b() { cout&lt;&lt;"\nIn function b() , c = "&lt;&lt;c(); } </code></pre> <p><br><strong>driver.cpp</strong><br/></p> <pre><code>#include&lt;iostream&gt; using namespace std; void a(); void b(); int c(); int main() { b(); a(); a(); b(); cout&lt;&lt;"\nIn function main() = , c "&lt;&lt; c(); cout&lt;&lt;"\n"; } </code></pre> <p><strong>output 1 :-</strong> </p> <pre><code>when compiles as follows :- bash#&gt;g++ -c test1.cpp bash#&gt;g++ -c test2.cpp bash#&gt;g++ -c driver.cpp bash#&gt;g++ -o out driver.o test1.o test2.o bash#&gt;./out In function b() , c = 1 In function a() , c = 2 In function a() , c = 3 In function b() , c = 4 IN main() , c = 5 </code></pre> <p>In above output , compiler is considering c() defined in test1.cpp</p> <p><strong>output 2:-</strong> changing sequence of test1.o and test2.o while linking.</p> <pre><code>bash#&gt;g++ -o out driver.o test2.o test1.o In function b() , c = 13 In function a() , c = 14 In function a() , c = 15 In function b() , c = 16 IN main() , c = 17 </code></pre> <p>In above output , compiler is considering c() defined in test2.cpp</p> <p>I was perplexed when i made minor changes in the code , which are as follows :- <br>1) if I do not call c() in function a() [test1.cpp] and c() in funciton b()[test2.cpp]<br/></p> <pre><code>//test1.cpp changes void a() { cout&lt;&lt;"\nIn function a() , c = "; // not calling c() } //test2.cpp changes void b() { cout&lt;&lt;"\nIn function b() , c = "; // not calling c() } </code></pre> <p>I get following error while linking :-</p> <pre><code>bash#&gt;g++ -o out driver.o test1.o test2.o driver.o: In function `main': driver.cpp:(.text+0x1f): undefined reference to `c()' collect2: ld returned 1 exit status </code></pre> <p>2) If i call c() in any one of the file i.e either in test1.cpp or in test2.cpp , then i wont get linker error.</p> <p>Could anyone please help me in understanding this behaviour.</p> <p>Thanks in advance.</p>
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    1. COTo avoid linker issues, move your `inline` functions to the header files.
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      1. USThomas Matthews
 

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