terminfo参数化字符串%d的编码行为

我正在尝试理解terminfo的参数化字符串解析器中%d编码的行为。相关的手册页在这里，并指出-

%[[:]flags][width[.precision]][doxXs]
as  in  printf, flags are [-+#] and space.  Use a ":" to allow the
next character to be a "-" flag, avoiding interpreting "%-" as  an
operator.

但没有说明从哪里打印值以及如何处理边缘案例。它们是来自堆栈还是来自传递给参数化字符串的参数？此外，当传递了额外的参数(参数化字符串中的%d不相等)或存在额外的%d(参数化的字符串不正确？)时会发生什么？这种未定义的行为或实现是在某个地方定义的还是由定义定义的？

我试图通过手动写入一些有效和无效的字符串并验证输出来检查一些情况，但到目前为止一切都有点不一致，所以我在这里看不到模式-

#include <iostream>
#include <curses.h>
#include <term.h>
using namespace std;
int main() {
// single %d prints single argument
// => 2
auto res = tparm("[%d]", 2);
cout << res << endl;
// single %d prints single argument and ignores additional
// => 2
res = tparm("[%d]", 2, 3, 4);
cout << res << endl;
// multiple %d prints 0 for absent additional arguments
// => 2-0-0-0
res = tparm("[%d-%d-%d-%d]", 2);
cout << res << endl;
// multiple %d prints with equal number of arguments prints
// first two correctly and rest 0
// => 2-3-0-0-0
res = tparm("[%d-%d-%d-%d-%d]", 2,3,4,5,6);
cout << res << endl;
// single value pushed to stack prints from stack
// => 2
res = tparm("[%p1%d]", 2);
cout << res << endl;
// single value pushed to stack prints from stack and ignores extra arguments
// => 2
res = tparm("[%p1%d]", 2,3,4);
cout << res << endl;
// single value pushed to stack prints from stack and additional prints are 0
// if no arguments are provided
// => 2-0-0
res = tparm("[%p1%d-%d-%d]", 2);
cout << res << endl;
// single value pushed to stack prints from stack and additional prints 0
// even if equal arguments are provided
// => 2-0-0
res = tparm("[%p1%d-%d-%d]", 2,3,4);
cout << res << endl;
// single value pushed to stack prints from stack after pop()?
// => 100-<garbage>
res = tparm("[%p1%d-%c]", 100);
cout << res << endl;
// pushed to stack via {} and equal arguments provided, prints all
// => 2-1-100-200
res = tparm("[%{1}%{2}%d-%d-%d-%d]", 100, 200);
cout << res << endl;
// pushed to stack via {} and %p1 equal arguments provided
// prints only stack and rest 0
// => 100-2-1-0
res = tparm("[%{1}%{2}%p1%d-%d-%d-%d]", 100, 200);
cout << res << endl;
}