Converting from int to char where overflow errors are concerned

Question

Isn't it a generally a bad idea to convert from a larger integral type to a smaller signed if there is any possibility that overflow errors could occur? I was surprised by this code in C++ Primer (17.5.2) demonstrating low-level IO operations:

int ch;

while((ch = cin.get()) != EOF)
    cout.put(ch); //overflow could occur here

Given that

cin.get() converts the character it obtains to unsigned char, and then to int. So ch will be in the range 0-255 (exluding EOF). All is good.
But then in the put(ch) expression ch gets converted back to char. If char is signed then any value of ch from 128-255 is going to cause an overflow, surely?

Would such code be generally bad practice if I'm expecting something outside of ordinary input 0-127, since there are no guarantees how overflow is treated?

cin.get guarantees that if the return value isn't EOF then it's representable as unsigned char. — Kerrek SB
– Kerrek SB, Commented Dec 17, 2015 at 15:08
@KerrekSB I thought that was what I said. Maybe I should've been clearer, my concern is with the cout.put(ch) statement. — SergeantPenguin
– SergeantPenguin, Commented Dec 17, 2015 at 15:16
Ah, yes. As a rule, all the IO functions consider char values to be converted from unsigned char, so this is fine. Note that the conversion is unspecified and not "overflow". — Kerrek SB
– Kerrek SB, Commented Dec 17, 2015 at 15:48

Kahler · Accepted Answer · 2015-12-17 15:24:07Z

1

There are rules for integer demotion.

When a long integer is cast to a short, or a short is cast to a char, the least-significant bytes are retained.

As seen in: https://msdn.microsoft.com/en-us/library/0eex498h.aspx

So the least significant byte of ch will be retained. All good.

answered Dec 17, 2015 at 15:24

Kahler

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Arash Hatami · Accepted Answer · 2015-12-17 15:27:45Z

0

Use itoa, if you want to convert the integer into a null-terminated string which would represent it.

char *  itoa ( int value, char * str, int base );

or you can convert it to a string , then char :

std::string tostr (int x){
    std::stringstream str;
    str << x;
    return str.str();}

convert string to char

string fname;
char *lname;
lname = new char[fname.lenght() + 1];
strcpy(f, lname.c_str());

if you see "Secure Error" disable it with #Pragma

edited Dec 17, 2015 at 15:27

answered Dec 17, 2015 at 15:21

Arash Hatami

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