堆栈：无法将几个字符压入数组 [英] Stack : not able to push a couple of characters into array

问题描述

我有一个代码可以使用数组实现堆栈，这里是完整的代码：这里

I have a code to implement a stack using array, here the complete code : here

这种情况就是为什么我不能推多个字符，而
只推一个字符？
，但是我一直在单击一些变量，这些变量使用 struct 初始化，以便她将
的某些字符以数组的形式推送给

The case is why I cannot to push more than one character, but only one character? but I've been clicking some variables are initialized using a struct for her push some characters in the form of an array:

struct stackMhs {
char nama[10];
char npm[10];
char telp[10];
int size;
};

struct stackMhs stackMhsbaru;

这是 push（）函数参数，该参数将成为函数 main（）中的数据内容：

This is the push() function with a parameter which will be the contents of data in the function main():

void push(char nm, char np, char tel) {
if(stackMhsbaru.size != 10) {
stackMhsbaru.nama[stackMhsbaru.size + 1] = nm;
stackMhsbaru.npm[stackMhsbaru.size + 1] = np;
stackMhsbaru.telp[stackMhsbaru.size + 1] = tel;
stackMhsbaru.size++;
}
else {
printf("stack is full!");
}
}

问题是我使用'在 push（）函数中仅一个字符，例如 push（'a'，'b '，'c'）; 在编译时没有错误，但是当我使用 如 push（ aa ， bb， cc）; 发生编译错误：

The problem is when I use ' to fill data is only one character at the push() function like push('a','b','c'); when at compile no errors, but when I use " such as push("aa","bb","cc"); when at compile error occurs:

main.c: In function 'main':
main.c:60:6: warning: passing argument 1 of 'push' makes integer from pointer without a cast [-Wint-conversion] 
 push("aa", "bb", "cc"); 
      ^ 
main.c:23:6: note: expected 'char' but argument is of type 'char *'
void push(char nm, char np, char tel) {
     ^
main.c:60:12: warning: passing argument 2 of 'push' makes integer from pointer without a cast [-Wint-conversion]
push("aa", "bb", "cc");
           ^
main.c:23:6: note: expected 'char' but argument is of type 'char *'
void push(char nm, char np, char tel) {
     ^
main.c:60:18: warning: passing argument 3 of 'push' makes integer from pointer without a cast [-Wint-conversion]
push("aa", "bb", "cc");
                 ^
main.c:23:6: note: expected 'char' but argument is of type 'char *'
void push(char nm, char np, char tel) {
     ^ 

我的问题是：有解决方案吗？

My question is : any solution ?

推荐答案

在C语言中，''内的任何字符都是（使用 char 声明，并且 内的任何内容都表示一个字符串，该字符串是 char s。

In C, anything inside a '' is a character (declared using char) and anything inside "" represents a string which is a null terminated array of chars.

不能将 char s数组分配给单个 char 变量，因此您会看到警告。

You cannot assign an array of chars to a single char variable, hence you see the warnings.

警告的解释：

警告：传递'push'的参数1会使指针的整数不进行强制转换

出于解释的原因，假定编译器在显示 int 时表示 char ，然后基本上是在抱怨您试图将 char 数组类型分配给 char ，而没有明确告诉编译器您想要这样做。

For explanation sake, assume that the compiler means char when it says int, then basically it is complaining that you are trying to assign a char array type to a char without explicitly telling compiler that you want to do so.

正确的操作方式：

逐个字符地将字符串传递给 push 函数在循环内调用。

Pass the string character by character to the push function called inside a loop.

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