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c - 由于它们是字符串,我将如何添加值?

转载 作者:行者123 更新时间:2023-11-30 19:16:29 26 4
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我已经尝试过使用atoi,然后将它们切换回字符串进行推送,我正在尝试为类制作一个 rpn 计算器,因此推送、弹出、查找和堆栈结构就是接下来需要的是,但我无法让它添加整数值。

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

struct stack
{
const char * data;
struct stack *bottom;
};

struct stack * push(struct stack *stk,const char * x)
{
struct stack *nstk = (struct stack *)malloc(sizeof(struct stack));

nstk->data = x;
nstk->bottom = stk;

return nstk;
}


const char * peek(struct stack *stk)
{
if(stk -> data)
return stk->data;
else
return("Stack is empty");
}

struct stack *pop(struct stack *stk)
{
struct stack *tmp;

tmp = stk->bottom;
stk->bottom = NULL;
free(stk);

return tmp;
}

FILE * input_from_args(int argc,const char *argv[])
{
if(strcmp(argv[1],"-e") != 0 && strcmp(argv[1],"-c") != 0 && strcmp(argv[1],"-g") != 0)
{
printf("Option %s is not supported \n", argv[1]);
exit(0);
}

else
{
return stdin;
}
}
void evaluate(struct stack * equation)
{
int op;
int op2;
int ans;

if(strcmp("A",equation->data) == 0)
{
op = (int)pop(equation)-> data;
op2 = (int)pop(equation)-> data;
ans = op + op2;
printf("%i",ans);
}
}

void convert(struct stack * equation)
{
}

void other (struct stack * equation)
{

}


int main(int argc,const char *argv[])
{
FILE *src = input_from_args(argc, argv);

if (src == NULL)
{
printf("%s", "Invalid Source");
exit(EXIT_FAILURE);
}

struct stack * equation = NULL;

int i;
for(i=2; i <= argc; i++)
{
equation = push(equation,argv[i]);
}

if(strcmp(argv[1],"-e") == 0)
{
evaluate(equation);
}
else if(strcmp(argv[1],"-c") == 0)
{
convert(equation);
}
else if(strcmp(argv[1],"-g") == 0)
{
other(equation);
}

return EXIT_SUCCESS;
}

这就是到目前为止我所拥有的一切,如果您注意到任何其他问题也没关系,但我真正想知道的是如何使用此数据结构评估后缀方程(输入示例为 -e 2 2 A 5 X)。

最佳答案

希望对您有所帮助。

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

typedef int DataType;

DataType cnv(const char *s){
return atoi(s);
}

#define PRN_DATA "%d"

typedef struct stack {
DataType data;
struct stack *bottom;
} Stack;

void push(Stack **stk, DataType x){
struct stack *nstk = malloc(sizeof(Stack));

nstk->data = x;
nstk->bottom = *stk;

*stk = nstk;
}

int empty(Stack *stk){
return stk == NULL;
}

DataType pop(Stack **stk){
struct stack *tmp = *stk;
if(empty(*stk)){
printf("empty stack\n");
exit(EXIT_FAILURE);
}
DataType ret = tmp->data;

*stk = (*stk)->bottom;
free(tmp);

return ret;
}

char input_from_args(int *argc, const char ***argv){
if( *argc < 2 || (*argv)[1][0] != '-' || (*argv)[1][1] == '\0'){
printf("Option is not specified\n");
exit(EXIT_FAILURE);
}
char op = (*argv)[1][1];
if((*argv)[1][2] != '\0' || op != 'e' && op != 'c' && op != 'g'){
printf("Option %s is not supported \n", (*argv)[1]);
exit(EXIT_FAILURE);
}
*argv = &(*argv)[2];
*argc -= 2;
return op;
}

void evaluate(int argc, const char **argv){
struct stack *s = NULL;
int i;
DataType v1, v2;
for(i = 0; i < argc; ++i){
switch(*argv[i]){
case 'A':
v2 = pop(&s);
v1 = pop(&s);
push(&s, v1 + v2);
break;
case 'S':
v2 = pop(&s);
v1 = pop(&s);
push(&s, v1 - v2);
break;
case 'X':
v2 = pop(&s);
v1 = pop(&s);
push(&s, v1 * v2);
break;
case 'D':
v2 = pop(&s);
v1 = pop(&s);
push(&s, v1 / v2);
break;
default:
push(&s, cnv(argv[i]));
}
}
printf(PRN_DATA "\n", pop(&s));
if(!empty(s)){
printf("data remains in the stack\n");
exit(EXIT_FAILURE);
}
}

void convert(int argc, const char **argv){
}

void other (int argc, const char **argv){
}

int main(int argc, const char **argv){
switch(input_from_args(&argc, &argv)){
case 'e':
evaluate(argc, argv);
break;
case 'g':
other(argc, argv);
break;
case 'c':
convert(argc, argv);
break;
}

return EXIT_SUCCESS;
}

关于c - 由于它们是字符串,我将如何添加值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30132408/

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