在C中如何将局部函数变量值传递给另一个函数并打印值而不会出现错误

#include <stdio.h> 
#include <stdlib.h>

int give_options();

int give_options(){

    printf("===== MENU 1 =====\n");

    printf("> store new customer :\t\tenter 1\n");
    printf("> view a customer :\t\tenter 2\n");
    printf("> view all customers :\t\tenter 3\n");
    printf("> exit program :\t\tenter 4\n ");   


    int ipt = malloc(sizeof(int));
    scanf("%d",&ipt);

    return ipt;
}


int main(){
    int dec = give_options();

    printf("decision is: %d", dec);     


    getchar();
}

Ive最近开始在Ubuntu中使用C编写代码，试图将值返回一个函数的局部变量，并将其传递给另一个函数。我已经读到，由于将局部变量分配给堆栈，因此函数返回后该值将不再存在，并且我必须使用malloc在堆中分配内存。

当我编译时，我得到以下警告：

initialization makes integer from pointer without a cast
[-Wint-conversion] int ipt = malloc(sizeof(int));

当我将其调整为int ipt =（int）malloc（sizeof（int））; 我得到：

warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
int ipt = (int)malloc(sizeof(int));

正确的做法是什么？

自从我切换到Ubuntu编码以来，我也遇到了scanf的麻烦。我尝试了另一种方法，尝试使用指针：

#include <stdio.h> 
#include <stdlib.h>

int* give_options();

int* give_options(){

    printf("===== MENU 1 =====\n");

    printf("> store new customer :\t\tenter 1\n");
    printf("> view a customer :\t\tenter 2\n");
    printf("> view all customers :\t\tenter 3\n");
    printf("> exit program :\t\tenter 4\n ");   

    int* ipt; 
    ipt = malloc(sizeof(int));
    scanf("d",&ipt);

    return ipt;
}


int main(){
    int* dec = give_options();

    printf("decision is: %d", dec); 

    getchar();
}

编译这个给了我以下错误：

warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int **’ [-Wformat=]
  scanf("%d",&ipt);

warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]
  printf("decision is: %d", dec);

哪种方法正确，哪种方法有效？