C language uses stack to calculate infix expressions

cThe language uses a stack to implement computing infix expressions

解决这个问题需要两步：

1. Change infix to postfix（The suffix type is more convenient for writing algorithms）
2. 计算后缀式.

Because it has to meet multiple digits and decimals,So strings are used.

typedef struct
{
    
	char c[100];
}X;
typedef struct
{
    
	X c[MAXSIZE];	
	int top;
}Sqstack;

Some functions on the stack will not be written,应该都知道.

The following directly changes the suffix code to the infix code（有注释）.

void Change(Sqstack S, X a[], X b[])//Turns an infix expression into a postfix expression,ais the infix form of the input,bis the converted suffix.
{
    
	int i;
	for (i = 0; a[i].c[0] != '#'; i++)//遍历输入字符串,无非三种情况,遇到数字、括号（大中小）、运算符.
	{
    
		if (a[i].c[0] >= 48 && a[i].c[0] <= 57)//When you encounter numbers, just throw them away
		{
    
			b[n] = a[i];
			n++;
		}
		if (a[i].c[0] == '+' || a[i].c[0] == '-' || a[i].c[0] == '*' || a[i].c[0] == '/')//遇到运算符,If the priority is greater than the top operator, it is pushed onto the stack,Otherwise, spit the top element of the stack,再次比较,Until the priority is less than the top operator or the stack is empty,入栈.**（Parentheses have the lowest precedence）**
		{
    
			if (Isempty(S))
			{
    
				S = Push(S, a[i]);
			}
			else
			{
    
				int j;
				for (j = S.top; j >= 0; j--)
				{
    
					if (Priority(a[i].c[0], S.c[j].c[0]))
					{
    
						S = Push(S, a[i]);
						break;
					}
					else
					{
    
						b[n] = S.c[j];
						n++;
						S = Pop(S);
					}
				}
				if (j == -1)
				{
    
					S = Push(S, a[i]);
				}
			}
		}
		if (a[i].c[0] == '(' || a[i].c[0] == '[' || a[i].c[0] == '{')//When the left parenthesis is encountered, it is directly pushed onto the stack
		{
    
			S = Push(S, a[i]);
		}
		if (a[i].c[0] == ')')//遇到右括号,Spit out the operators in the stack one by one.
		{
    

			if (a[i+1].c[0] == '*' || a[i+1].c[0] == '/')//This sentence is the solution（4+3）*2 这种情况.
			{
    
				for (int j = S.top; j >= 0; j--)
				{
    
					if (S.c[j].c[0] == '(')
					{
    
						S = Pop(S);
						break;
					}
					else
					{
    
						b[n] = S.c[S.top];
						n++;
						S = Pop(S);
					}
				}
			}
			else
			{
    
				for (int j = S.top; j >= 0; j--)
				{
    
					if (S.c[j].c[0] == '(')
					{
    
						S = Pop(S);
						for (int k = j - 1; S.c[k].c[0] != '['&&S.c[k].c[0] != '{'&&k >= 0; k--)
						{
    
							b[n] = S.c[k];
							n++;
							S = Pop(S);
						}
						break;
					}
					else
					{
    
						b[n] = S.c[S.top];
						n++;
						S = Pop(S);
					}
				}
			}
		}
		if ( a[i].c[0] == ']' )
		{
    
			if (a[i + 1].c[0] == '*' || a[i + 1].c[0] == '/')
			{
    
				for (int j = S.top; j >= 0; j--)
				{
    
					if (S.c[j].c[0] == '[')
					{
    
						S = Pop(S);					
						break;
					}
					else
					{
    
						b[n] = S.c[S.top];
						n++;
						S = Pop(S);
					}
				}
			}
			else
			{
     
				for (int j = S.top; j >= 0; j--)
				{
    
					if ( S.c[j].c[0] == '[')
					{
    
						S = Pop(S);
						for (int k = j - 1; S.c[k].c[0] != '{'&&k >= 0; k--)
						{
    
							b[n] = S.c[k];
							n++;
							S = Pop(S);
						}
						break;
					}
					else
					{
    
						b[n] = S.c[S.top];
						n++;
						S = Pop(S);
					}
				}
			}
		}
		if ( a[i].c[0] == '}')
		{
    
			if (a[i + 1].c[0] == '*' || a[i + 1].c[0] == '/')
			{
    
				for (int j = S.top; j >= 0; j--)
				{
    
					if (S.c[j].c[0] == '{')
					{
    
						S = Pop(S);						
						break;
					}
					else
					{
    
						b[n] = S.c[S.top];
						n++;
						S = Pop(S);
					}
				}
			}
			else
			{
    
				for (int j = S.top; j >= 0; j--)
				{
    
					if (S.c[j].c[0] == '{')
					{
    
						S = Pop(S);
						for (int k = j - 1; k >= 0; k--)
						{
    
							b[n] = S.c[k];
							n++;
							S = Pop(S);
						}
						break;
					}
					else
					{
    
						b[n] = S.c[S.top];
						n++;
						S = Pop(S);
					}
				}
			}
		}
	}
	for (int k = S.top; k >= 0; k--)
	{
    
		b[n] = S.c[k];
		n++;
		S = Pop(S);
	}
}

Then there is the algorithm of the operation.（这个就比较简单了）

void Operate(Sqstack S, X b[])//计算表达式
{
    
	int i;
	for (i = 0; i < n; i++)
	{
    
		if (b[i].c[0] >= 48 && b[i].c[0] <= 57)
		{
    
			S = Push(S, b[i]);
		}
		else//Know that the operator is encountered.Pop the first two elements from the top of the stack, compute the result, and push the result onto the stack.
		{
    
			X m, n, x;
			strcpy(m.c, S.c[S.top].c);
			S = Pop(S);
			strcpy(n.c, S.c[S.top].c);
			S = Pop(S);
			Calculate(m.c, b[i].c[0], n.c, x.c);
			S = Push(S,x);
		}
	}
	X y;
	y = S.c[S.top];
	double t = atof(y.c);
	printf("%.2f", t);
}

Another detail is the conversion of string and floating-point data（当然intTypes are also converted,But also count the decimals,So choose floating point）

void Calculate(char b[], char c, char a[],char x[])//计算
{
    
	double m = atof(a);
	double n = atof(b);
	int dec_pl, sign;
	switch (c)
	{
    
		case '+':
		{
    
			double z = m + n;
			_gcvt(z, 8, x);
			break;
		}
		case '-':
		{
    
			double z = m - n;
			_gcvt(z, 8, x);
			break;
		}
		case '*':
		{
    
			double z = m * n;
			_gcvt(z, 8, x);
			break;
		}
		case '/':
		{
    
			double z = m / n;
			_gcvt(z, 8, x);
			break;
		}
	}
}

Some knowledge is quoted in https://blog.csdn.net/coder_dacyuan/article/details/79941743

新手上路,写的比较水,谢谢观看.嘻嘻.