Done. First task #1
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yegorjke
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yegorjke
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remove unnecessary couts and endls. make yours "print" function working. fix other comments
| auto adglob{34.5}; | ||
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| // The function doesn't work because global variables sent with copy of value | ||
| // template <typename T> inline void print_vars(string name, T var, T * pvar, T & rvar) |
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I took a look at this function and it seems me overhead. I don't understand why you need to have all these passed params (I followed that these are a value, pointer, and reference of the same variable, right?), but if consider it as a real function of your library, do you really always have them in present on the stack? If not, you will have to pass them because the function demands (even so, print_vars("var", var, &var, var)).
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If you don't want to pass params as values then consider passing them as [const] reference.
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If I were you I would like to have the function which takes 1 required argument var, it is a variable being printed, and the optional one name or description. The function itself will perform printing the value, the address in memory, and the sizeof (and type of variable (see typeid) but it is optional). Some kind of this:
template<typename T>
void print_variable_info(const T& var, const std::string& name = "") {
std::string prefix = (name.empty()) ? "" : ("[" + name + "] ");
std::cout << prefix << "Value: " << var << "\n";
std::cout << prefix << "Address: " << &var << "\n"; // Q: why it won't print address of var with type "char"? what is needed to do?
std::cout << prefix << "Type: " << typeid(var).name() << "\n";
std::cout << prefix << "Sizeof: " << sizeof(var) << std::endl;
}It doesn't matter what you will pass to this function. If it confuses you how it will work with a pointer, then I would say that a pointer is a variable as well, but the value of a pointer is address (and the pointer has its own address as well as a common variable)
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Done. The source code is changed to use template function.
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std::cout << prefix << "Address: " << &var << "\n"; doesn't work because the compiler interpret &char variable as c-style string and tries to print whole string instead of address of pointer. To print address of char pointer we need to use static_cast<const void *>.
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| cout << "---------------------------------------------------------------------" << endl; | ||
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| cout << "Value of iglob: " << iglob << endl; |
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remake it with the proposed function above
| //ERROR: error: invalid conversion from ‘const char*’ to ‘int’ [-fpermissive] | ||
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| return 0; | ||
| } No newline at end of file |
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add to habits leaving an extra newline at the EOF.
| cout << "c - const" << endl; | ||
| cout << "s - static" << endl; | ||
| cout << "glob - global" << endl; | ||
| cout << "loc - local" << endl << endl; |
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a lot of "endl"s force the IO buffer to flush. it costs the time of your [future] programs. it is useful to flush the buffer when you are sure that it must be done. consider of "\n" in some places.
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Got it. Thanks. I will use the advice in the future.
| unsigned char * puchar{&ucloc}; | ||
| cout << "*puchar <- ucloc " << endl << "*puchar=" << int (*puchar) << endl; | ||
| *puchar=*puchar +1; | ||
| cout << "increment *puchar " << endl << "*puchar=" << int (*puchar) << endl; |
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Added print of ucloc after pointer value increment.
| float floc{12.34}; | ||
| double dloc{12.3456}; | ||
| char cloc{'l'}; | ||
| unsigned char ucloc{'k'}; |
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what if ucloc is assigned with a small integer (for example, 12)? what will be printed?
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Nothing will be printed until k>32. Because from 0 to 31 are not printable signs, 32 is space and from 33 are printable signs.
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Made corrections in accordance with the comments |
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Finish the first task.