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POSize of structure with a char, a double, an int and a t
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copied!<p>When I run only the code fragment</p> <pre><code>int *t; std::cout << sizeof(char) << std::endl; std::cout << sizeof(double) << std::endl; std::cout << sizeof(int) << std::endl; std::cout << sizeof(t) << std::endl; </code></pre> <p>it gives me a result like this:</p> <pre><code>1 8 4 4 </code></pre> <p>Total: 17.</p> <p>But when I test sizeof struct which contains these data types it gives me 24, and I am confused. What are the additional 7 bytes?</p> <p>This is the code</p> <pre><code>#include <iostream> #include <stdio.h> struct struct_type{ int i; char ch; int *p; double d; } s; int main(){ int *t; //std::cout << sizeof(char) <<std::endl; //std::cout << sizeof(double) <<std::endl; //std::cout << sizeof(int) <<std::endl; //std::cout << sizeof(t) <<std::endl; printf("s_type is %d byes long",sizeof(struct struct_type)); return 0; } </code></pre> <p>:EDIT</p> <p>I have updated my code like this</p> <pre><code>#include <iostream> #include <stdio.h> struct struct_type{ double d_attribute; int i__attribute__(int(packed)); int * p__attribute_(int(packed));; char ch; } s; int main(){ int *t; //std::cout<<sizeof(char)<<std::endl; //std::cout<<sizeof(double)<<std::endl; //std::cout<<sizeof(int)<<std::endl; //std::cout<<sizeof(t)<<std::endl; printf("s_type is %d bytes long",sizeof(s)); return 0; } </code></pre> <p>and now it shows me 16 bytes. Is it good, or have I lost some important bytes?</p>

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