Member Avatar for prushik

I would like to know once and for all how to convert a double to a const char*

I've tried it a few times, but have never succeeded, I know it can be done, but I haven't figured out how to yet. I've tried some different things. Sometimes it won't compile, and sometimes the program crashes, and other times it ends up as an empty string.

  • 6 Contributors
  • 5 Replies
  • 2K Views
  • 1 Year Discussion Span
  • Latest Post Latest Post by Seta00

Dani AI

Generated

You cannot cast a double to const char*. A cast only reinterprets bytes; it does not turn 12.35 into the text "12.35". As and hint, you must format the value into characters, then pass a pointer to that character storage.

C-style, fixed buffer (bounds-checked):

#include <cstdio>

char buf[64];
double x = 12.35;
int n = std::snprintf(buf, sizeof buf, "%.3f", x);
if (n < 0 || static_cast<size_t>(n) >= sizeof buf) {
    // handle error or truncation
}
// use buf as const char*

Prefer snprintf over sprintf because it limits writes to the buffer size. See the <cstdio> signatures for sprintf/snprintf. (en.cppreference.com)

Modern C++ options that still give you const char* when needed:

// C++11: simple, but fixed formatting
std::string s = std::to_string(x);   // prints 6 digits after the decimal by default
const char* p = s.c_str();           // valid while 's' is alive and unmodified

std::to_string follows printf-style defaults (six fractional digits; uses the C locale), which may or may not be what you want. (en.cppreference.com) The pointer from c_str() stays valid until the string is modified or destroyed, so do not return it from a function unless the std::string outlives that use. (my.eng.utah.edu)

If you want precise control without allocations, use C++17 to_chars into your own buffer:

#include <charconv>

char buf2[64];
auto [ptr, ec] = std::to_chars(std::begin(buf2), std::end(buf2),
                               x, std::chars_format::fixed, 3);
if (!ec) { *ptr = '\0'; /* use buf2 */ }

to_chars is locale-independent, non-allocating, and does not append a null terminator for you. (en.cppreference.com)

Bonus: In C++20, std::format gives printf-like control but returns a std::string; call c_str() if a const char* is required. (en.cppreference.com)

Recommended Answers

All 5 Replies

Member Avatar for amt_muk

U can try sprintf().

Amit

Member Avatar for Ancient Dragon

depends on how you want to use it. If you want to pass a pointer to a double as a parameter to some function that takes char* then typecasting might work, depending on what that function is going to do with it. Normally, however, you will have to convert it storing the results in either a character array or a std::string object.

Two ways come to mind: sprintf() will convert and store in a character array, while std::stringstream will convert and store in std::string object.

Member Avatar for Ptolemy

>I would like to know once and for all how to convert a double to a const char*
Casting doesn't work. You're not converting the double to a string with a cast, you're telling the compiler to use the binary representation of the double as characters in a string. For a value like 12.35, that's not going to give you a string like "12.35". You need to take the printed value (like if you write the double using cout) and store it in a string. That's really really painful to do manually with floating-point, so you should use sprintf or stringstream like Ancient Dragon said.

Member Avatar for Seta00


____________________
Le Parkour LifeStyle

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.