The code works perfectly when I substitute char with string.why so?

Question

The below code works perfectly with strings but with char it gives segmentation fault.

#include<iostream>
using namespace std;

class salary
{
public:
    int empno;
    float inctax;
    float netsal;
    int gross;
    short int age;
    char name[50];

    salary(){
        empno=0;
        gross=0;
        age=0;
        strcpy(name,'\0');
    }

    salary(int empn,int gros,short int ag,char nam[]){
        empno=empn;
        gross=gros;
        age=ag;
        strcpy(name,nam);
    }

    void calc(){
        inctax=0.0;
        if(gross>1000000)
            inctax=0.3*gross;
        else if(gross>=500000 && gross<=1000000)
            inctax=0.2*gross;
        else if(gross>=250000 && gross<500000)
            inctax=0.1*gross;
        else
            inctax=0.0;
        netsal=gross-inctax;
        cout<<"inctax"<<inctax;
        cout<<"net sal"<<netsal;
    }
};

int main(){
    salary *r=new salary();
    salary *r1=new salary(112,500000,21,"Arnab");
    r1->calc();
    return 0;
}

Please indent your code. It is impossible to read in the current format. — abhishek_naik, Jun 30 '16 at 03:48
"works perfectly with strings but with char it gives segmentation fault." What do you mean by that? Which strings and which char? — Ken Y-N, Jun 30 '16 at 03:51
`strcpy(name,'\0');` is an error, either use `strcpy(name,"\0");` or `name[0] = '\0';` — kmdreko, Jun 30 '16 at 03:56
You are more likely to get help here if you can simplify your question and ask it as clearly as possible. For example, I'm not sure what you're asking but I'm pretty confident the `calc()` function doesn't need to be here. — Dan R, Jun 30 '16 at 04:02
Can you point to exactly where this substitution is occurring? At the moment it is not clear. — Paul Rooney, Jun 30 '16 at 04:05
What is it to replace with `char` to generate the same problem? — Dean Seo, Jun 30 '16 at 04:19
My suspicion is that the OP means `strcpy(name,"");` works, but `strcpy(name,'\0');` doesn't. — Ken Y-N, Jun 30 '16 at 05:03

score 3 · Accepted Answer · answered Jun 30 '16 at 04:25

3

I agree with vu1p3n0x's comment, the problem is in the default constructor.

strcpy(name, '\0'); // This is wrong!

strcpy() takes two char arrays as arguments, but a character itself was passed as the second argument in your code. The syntax of strcpy() is:

char * strcpy ( char * destination, const char * source );

In order to create null string using char arrays, you should probably use,

strcpy(name, "\0");
     /*or*/ 
name[0] = '\0';

More on copying string, Reference and tutorialspoint

answered Jun 30 '16 at 04:25

Praveen Kishore

436
1
3
14

can you explain why single quotes for normal char usage and double quotes in strcpy? – Arnab Mitra Jun 30 '16 at 09:16
Have a look at this, [Thread](http://stackoverflow.com/questions/3683602/single-quotes-vs-double-quotes-in-c). Explains the difference – Praveen Kishore Jun 30 '16 at 13:39

score 1 · Answer 2 · answered Jun 30 '16 at 16:46

Your main problem is that your compiler isn't telling you about the obvious errors in the code. If you have correctly told it to report errors, then you perhaps need a better compiler. Here's the output I get when I compile:

g++ -std=c++14 -fPIC -g -Wall -Wextra -Wwrite-strings -Wno-parentheses     38113648.cpp   -o 38113648
38113648.cpp: In constructor ‘salary::salary()’:
38113648.cpp:18:25: error: ‘strcpy’ was not declared in this scope
         strcpy(name,'\0');
                         ^
38113648.cpp: In constructor ‘salary::salary(int, int, short int, char*)’:
38113648.cpp:25:24: error: ‘strcpy’ was not declared in this scope
         strcpy(name,nam);
                        ^
38113648.cpp: In function ‘int main()’:
38113648.cpp:46:48: warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
     salary *r1=new salary(112,500000,21,"Arnab");
                                                ^
38113648.cpp:45:13: warning: unused variable ‘r’ [-Wunused-variable]
     salary *r=new salary();
             ^

When I add

#include <cstring>

and change the constructor to take char const[], I get

g++ -std=c++14 -fPIC -g -Wall -Wextra -Wwrite-strings -Wno-parentheses     38113648.cpp   -o 38113648
38113648.cpp: In constructor ‘salary::salary()’:
38113648.cpp:19:25: warning: null argument where non-null required (argument 2) [-Wnonnull]
         strcpy(name,'\0');
                     ^

Obviously you meant that to be "\0". Better still, provide an initializer:

salary()
    : empno{0},
      gross{0},
      age{0},
      name{0}
{
}

salary(int empno, int gross, short int age, char const name[])
    : empno{empno},
      gross{gross},
      age{age}
{
    strcpy(this->name, name);
}

(I've also given the formal parameters more meaningful names, as this often forms the documentation of the constructor).

Adding -Weffc++ may also be worthwhile - in this case it warns that you don't initialize inctax or netsal in the constructors. If you're happy to have part-initialized objects, you'll obviously need to use Valgrind to check that these values are indeed set before they are used, because it can't be determined statically.

The code works perfectly when I substitute char with string.why so?

2 Answers2