Can the minimisation operation be seen from a programming language perspective?

Question

If $f$ is a total function $\mathbb N^k\to\mathbb N$, and $g$ is a total function $\mathbb N^{k+2}\to\mathbb N$, then we say that $h:\mathbb N^{k+1}\to\mathbb N$ is definable by primitive recursion from $f$ and $g$ if for all $x_1,\dots,x_k\in\mathbb N$ and $y\in\mathbb N$, \begin{align} h(x_1,\dots,x_k,0)&=f(x_1,\dots,x_k), \\ h(x_1,\dots,x_k,y')&=g(x_1,\dots,x_k,y,h(x_1,\dots,x_k,y)), \end{align} where $y'$ denotes the successor of $y$.

Primitive recursion can be seen as corresponding to a "for loop" in a programming language – the following pseudocode computes $h(x_1,\dots,x_k,y)$:

INPUT x1,...xk,y
value = f(x1,...,xk)
for i=0 to y:
    value = g(x1,...,xk,i,value)
RETURN value

I find such an explanation quite valuable, since it reinforces the idea that general recursive functions are ones that can be "computed" in a reasonable sense. Is there a similar way in which the minimisation operation can be understood? More specifically, does minimisation correspond to some kind of "loop" found in programming languages?

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For reference, the minimisation operation can be defined as follows:

If $f:\mathbb N^{k+1}\rightharpoonup \mathbb N$ is a partial function, then the minimisation $m:\mathbb N^k\rightharpoonup\mathbb N$ of $f$ is given by the following rule: if $f(x_1,\dots,x_k,t)$ is defined and $\neq0$ for all $t<y$, and $f(x_1,\dots,x_k,y)=0$, then $m(x_1,\dots,x_k):=y$; if there is no such $y$, then $m(x_1,\dots,x_k)$ is undefined.

Answer 1

Is there a similar way in which the minimisation operation can be understood? More specifically, does minimisation correspond to some kind of "loop" found in programming languages?

Yes, minimisation corresponds to possibly non-terminating loops. It increments t and searches for the first 0 value. When the function is undefined it means that it is looping forever.

INPUT x1,...xk
t = 1
value = f(x1,...,xk,t)
while value != 0:
     t = t + 1
     value = f(x1,...,xk,t)
RETURN t

Note that when f is undefined, it also would not terminate.