[SOLVED] unable to understand its execution, kindly help

[Hiroshi_S]

#include <stdio.h>
#include <string.h>
struct student
{
    char name[20];
}std;

char *fun(struct student *tempStd) // pointer
{
    strcpy(tempStd->name,"Thomas");
    return tempStd->name;
}

char *fun_one(struct student *tempStd)
{
    strcpy(tempStd->name,"Mike");
    return tempStd->name;
}
int main()
{
    strcpy(std.name,"Mike 2");
    printf("%s%s",fun(&std),fun_one(&std));
    return 0;
}

output : ThomasThomas

If I print as : printf("%s%s",fun_one(&std),fun(&std)) ,then the output is MikeMike.
Please help me to understand this concept.

 

[KlausST]

Hi,

I'm no C specialist, so see my post as an opinion. In doubt better rely on others..

In my opinion the struct "student" is useless. A struct mainly is used to join some variables to some kind of "dataset".
You may use the array "name" instead. If I'm not mistaken then an array (and a struct) is treated as a pointer and may be passed without "&" and "*".

But what the code is good for: I don't know. Maybe just for learning. Thus I think it's a good idea to give some context or link to a document where this code is from.

Klaus

 

[Hiroshi_S]

Yes, @KlausST , its for learning purpose for the understanding of pointer and structure.
ref link below. find question no.3.
C programming Structure and Union Aptitude Questions and Answers (includehelp.com)

 

[betwixt]

I agree, it is strange code but maybe only for educational purposes.
A struct containing only one item serves no useful purpose unless it is intended to be a base to add other things to it.

Brian.

 

[KlausST]

Hi,

You learn about structure... but not how it's meant to be. You don't learn it's true benefit.
As (funny) comparison: You learn to use a toothbrush ... for cleaning shoes.

Klaus

 

[Hiroshi_S]

Hi,

You learn about structure... but not how it's meant to be. You don't learn it's true benefit.
As (funny) comparison: You learn to use a toothbrush ... for cleaning shoes.

Klaus

HAHAHA.... its one of the strange code. Well lets not consider this as application oriented rather let us think why this code operates such way. I guess you guys might try to solve and came to output must be 'ThomasMike'. but its output is different. i am curious to know how it works.

 

[betwixt]

Remember that defining a struct just allocates space for it to hold it's component parts. It doesn't actually put anything in those spaces, at least in your example.

The code creates instances (copies) of the struct called 'std' and 'TempStd' which are the same design of structure but different copies of it. They both contain an entry called 'name' but it is a different entry in each instance of the struct. Note which is used when the output is printed.

Brian.

 

[FvM]

There are no copies, only pointers to instance "std". The functions are overwriting std.name through the pointers.

The excercise is revealing the order of function evaluation and and that the pointer content is printed after both functions have been evaluated. I'm not completely sure, but I believe that these details belong to the implementation dependend part of the specification. Respectively the observed behaviour must not necessarily be reproduced with different compilers.

 

[Hiroshi_S]

Yes, There is no copies, just the pointers.
And Thanks your support I was able to proceed this.

As you can see the code in the function 'fun' and 'fun_one' both of their return type is pointer pointing the location of the value.
as you can see in the 'printf' statement, those function are orderly executed from the left to right
First 'fun_one' which change the value of name as 'Mike' and return its pointer location not the data,
then the function 'fun' is executed where the value of name is changed to 'Thomas' and return the pointer location not the data. Now both the fun and fun_one points towards the same location but the current value of that pointer is 'Thomas'.
So as further execution of the printf goes, it will display two strings which at the pointer location which is 'Thomas'. so 'Thomas' is printed two times as 'ThomasThomas'.

similarly, if i change the position of function call in printf statement as

printf("%s%s",fun_one(&std),fun(&std));

then it will display as 'MikeMike', since 'fun_one' changes the name as 'Mike' so at the pointer location only Mike is stored thats why 'Mikemike' is printed.

huff.... i tried to debug that code by adding printf statement as below.

#include <stdio.h>
#include <string.h>

struct student{
    char name[20];
}std;

char *fun(struct student *tempstd){
    printf("1. %s \n",tempstd->name);
    strcpy(tempstd->name,"MIKE");
    printf("2. %s \n",tempstd->name);
    return tempstd->name;
}

char *fun_one(struct student *tempstd){
    printf("1. %s \n",tempstd->name);
    strcpy(tempstd->name,"THOMAS");
    printf("2. %s \n",tempstd->name);
    return tempstd->name;
}

int main()
{
    printf("Hello World\n");
    strcpy(std.name,"mike2p");
    printf("0.%s%s%s \n",std.name,fun_one(&std),fun(&std));
    return 0;
}
/*output
Hello World                                                                                                                  
1. mike2p                                                                                                                    
2. MIKE                                                                                                                      
1. MIKE                                                                                                                      
2. THOMAS                                                                                                                    
0.THOMASTHOMASTHOMAS*/

well, if you ask me what i learned, its that in printf statement, the exection order is from left to right. and then print it in the order we specified. and the pointer of struct pointers.
I also thought the pritf sequence changes form compiler to compiler, i tried various compiler GCC, Turboc, codeblock ide and online platform too. were all of them gave the same output.

and please correct me if i gone wrong some where.