Weird issue of variable function parameter

Question

I have a print function worked well, today I want add another parameter in this function, so I added it as first parameter, but all messages with extra parameters such as "%d", are all showing incorrectly, I searched for hours trying to understand why but got nothing useful, did I make anything wrong?

I tried to add "__cdecl", it didn't work either.

If I removed first parameter "int i", then everything is working properly. It's running on OS X 10.11 with Xcode 7.1.2 a console application.

void print(int i, const char* format_str, ...) {
    // get parameters
    va_list argptr;
    char buffer[1024] = {0};
    va_start(argptr, format_str);
    vsprintf(buffer, format_str, argptr);
    // log on screen
    std::cout << (char*)buffer << std::endl;
    va_end(argptr);
}

void call_print(const char* format_str, ...) {
    print(1, format_str);
}

int main(int argc, const char * argv[]) {
    // insert code here...
    int test = 2;
    call_print("test with %d", test);
    // print(1, "test with %d", test);
    return 0;
}

OK, I edited the code to make it run-able, it looks like the warped function is where the problem came out, how should I make it right??

If call print directly, it works properly, if using call_print, it will print "test with 1606416520", if removing the first parameter "int i", it works either....

where is `buffer` defined? assuming it is char `buffer[some-value]` then try `<< (char*)buffer` — slashmais, Feb 10 '16 at 12:05
buffer is pre-defined, and it's big enough, "<< (char*)buffer" is not working either :( — Jimmy liu, Feb 10 '16 at 12:09
look at this Q: https://stackoverflow.com/questions/2342162/stdstring-formatting-like-sprintf it may help you — slashmais, Feb 10 '16 at 12:17
@molbdnilo: i rarely use std::cout to do anything but output strings (if you look at my answer to the quoted question you'll see why) - as to your question, i don't know, what do you think? (rhetorical) — slashmais, Feb 10 '16 at 12:25

score 2 · Accepted Answer · answered Feb 10 '16 at 12:27

2

call_print is not passing the parameter 'test' on to print

skip call_print and call print directly

answered Feb 10 '16 at 12:27

slashmais

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what can I do if I really want to warp it. – Jimmy liu Feb 10 '16 at 12:30
I tried to put va_list and va_start in to call_print(), then pass the va_list to print(), it works but not looking nicely with this approach, I wonder if there is any way to make it work more elegantly. – Jimmy liu Feb 10 '16 at 12:37
@Jimmyliu: look at my answer here: https://stackoverflow.com/a/3742999/15161 - it may help – slashmais Feb 10 '16 at 12:40

Jimmy liu · Answer 2 · 2016-02-10T14:04:24.480

It seems I have 2 ways to make it work, one is C++11's variadic template and according to my project's situation I choose second way which is simpler by moving va_start() to warp function call_print() then pass va_list to print(), finalized code would like this:

void print(int i, const char* format_str, va_list& argptr) {
    char buffer[1024] = {0};
    vsprintf(buffer, format_str, argptr);
    // log on screen
    std::cout << (char*)buffer << std::endl;
}

void call_print(const char* format_str, ...) {
    // get parameters
    va_list argptr;
    va_start(argptr, format_str);
    print(1, format_str, argptr);
    va_end(argptr);
}

int main(int argc, const char * argv[]) {
    // insert code here...
    int test = 2;
    call_print("test with %d", test);
    // print(1, "test with %d", test);
    return 0;
}

And using C++11's variadic template would like this, need add "-std=c++11" in other C++ flags:

void print(int i, const char* format_str, ...) {
    // get parameters
    va_list argptr;
    char buffer[1024] = {0};
    va_start(argptr, format_str);
    vsprintf(buffer, format_str, argptr);
    // log on screen
    std::cout << (char*)buffer << std::endl;
    va_end(argptr);
}

template< typename... Args >
void call_print(const char* format_str, Args... args) {
    print(1, format_str, args...);
    //std::snprintf( nullptr, 0, format_str, args... );
}

int main(int argc, const char * argv[]) {
    // insert code here...
    int test = 2;
    call_print("test with %d", test);
    //print(1, "test with %d", test);
    return 0;
}

Weird issue of variable function parameter

2 Answers2