你好我想做这样的事情：
int a=1,b=2,c=3,w=4,result;
char d='+',e='-',f='/',g='*';
a d b g w =result;
printf("%d",result);
在此示例中，结果应为1 + 2 * 4 任何想法我怎么做？
答案 0 :(得分：0)
这可以做什么：a + b; a - b; a * b; a / b;
a / b不会给出正确答案，因为它正在进行整数除法。
正确的数学，煮咖啡，以及其他一切
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <ctype.h>
#include <string.h>
#include <errno.h>
enum operation {
INVALID,
CONTINUE, /* got just a number */
QUIT,
PLUS,
MINUS,
MULTIPLY,
DIVIDE
};
enum operation parse(char *expr, int *left, int *right);
int eval(const char *expr);
char *remove_spaces(char *expr);
int main() {
char expr[256];
FILE *fp = stdin;
char *rv;
int left, right;
enum operation oper = CONTINUE;
while (oper != QUIT) {
fputc(':', stdout);
rv = fgets(expr, sizeof(expr), fp);
if( rv == NULL ) {
if(feof(fp)) {
fprintf(stderr, "someone closed the input\n");
break;
}
if(ferror(fp)) {
fprintf(stderr, "something bad happened\n");
break;
}
}
oper = parse(expr, &left, & right);
/* this switch should be moved into an eval function
* which would deal with precedence of operations etc.
*/
switch(oper) {
case CONTINUE:
fprintf (stderr, "%d\n", left);
continue;
case INVALID:
fprintf(stderr, "> Don't know what to do with %-32s...\n", expr);
continue;
case QUIT:
fprintf(stderr, "> quitting\n");
break;
case PLUS:
fprintf(stdout, "%d\n", left + right);
break;
case MINUS:
fprintf(stdout, "%d\n", left - right);
break;
case MULTIPLY:
fprintf(stdout, "%d\n", left * right);
break;
case DIVIDE:
fprintf(stdout, "%d\n", left / right);
break;
}
}
return 0;
}
enum operation next_expr(char *expr, int *left, int *right)
{
enum operation oper;
char *nextbit = NULL;
*left = strtol(expr, &nextbit, 10);
if ((*left == 0 && errno == EINVAL) || nextbit == expr ) {
if(strcasecmp(expr, "quit") == 0)
return QUIT;
return INVALID;
}
if (*nextbit == 0 )
return CONTINUE;
switch(*nextbit) {
case '+': oper = PLUS; break;
case '-': oper = MINUS; break;
case '*': oper = MULTIPLY; break;
case '/': oper = DIVIDE; break;
default:
return INVALID;
}
expr = nextbit+1; /* move over to the next number */
*right = strtol(expr, &nextbit, 10);
if((*right == 0 && errno == EINVAL ) || nextbit == expr) {
return INVALID;
}
/* what if there are still more bits in next bit? */
return oper;
}
enum operation parse(char *expr, int *left, int *right)
{
enum operation suboperation;
remove_spaces(expr);
suboperation = next_expr(expr, left, right);
/* TODO: take care of the rest of the bits that are in the expr */
return suboperation;
}
char * remove_spaces(char *expr)
{
char *p = expr, *q = expr;
while(*q != 0) {
*p = *q;
q++;
if(!isspace(*p))
p++;
}
*p = 0;
return expr;
}
答案 1 :(得分：0)
您无法以C的方式执行动态算术。 您可以使用预处理器为编译时常量表达式播放有限的技巧，但对于通用解决方案，您需要一个表达式求值程序。
这是一个简单的评估器，支持4个基本操作员进行有限的健全性检查：
#include <ctype.h>
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
long long eval(const char *s) {
struct operand {
long long val;
int op, prec;
} stack[16], *sp, x;
for (sp = stack;;) {
x.val = strtoll(s, (char **)&s, 0);
while (isspace((unsigned char)*s))
s++;
switch (x.op = *s++) {
case '*':
case '/':
case '%': x.prec = 2; break;
case '+':
case '-': x.prec = 1; break;
default: x.prec = 0; x.op = 0; break;
}
while (sp > stack && x.prec <= sp[-1].prec) {
switch ((--sp)->op) {
case '*': x.val = sp->val * x.val; break;
case '/': x.val = sp->val / x.val; break;
case '%': x.val = sp->val % x.val; break;
case '+': x.val = sp->val + x.val; break;
case '-': x.val = sp->val - x.val; break;
}
}
if (!x.op)
return x.val;
if (sp == stack + sizeof(stack) / sizeof(stack[0])) {
errno = ERANGE; /* expression complexity error */
return 0;
}
*sp++ = x;
}
}
int main(int argc, char *argv[]) {
if (argc > 1) {
for (int i = 1; i < argc; i++) {
printf("%s -> %lld\n", argv[i], eval(argv[i]));
}
} else {
char buf[80];
int a = 1, b = 2, c = 3, w = 4, result;
for (int i = 0; i < 4*4*4; i++) {
snprintf(buf, sizeof buf, "%d %c %d %c %d %c %d",
a, "+-/*"[i / 16],
b, "+-/*"[i / 4 % 4],
c, "+-/*"[i % 4],
w);
result = eval(buf);
printf("%s -> %d\n", buf, result);
}
}
return 0;
}
输出：
1 + 2 + 3 + 4 -> 10
1 + 2 + 3 - 4 -> 2
1 + 2 + 3 / 4 -> 3
1 + 2 + 3 * 4 -> 15
1 + 2 - 3 + 4 -> 4
1 + 2 - 3 - 4 -> -4
1 + 2 - 3 / 4 -> 3
1 + 2 - 3 * 4 -> -9
1 + 2 / 3 + 4 -> 5
1 + 2 / 3 - 4 -> -3
1 + 2 / 3 / 4 -> 1
1 + 2 / 3 * 4 -> 1
1 + 2 * 3 + 4 -> 11
1 + 2 * 3 - 4 -> 3
1 + 2 * 3 / 4 -> 2
1 + 2 * 3 * 4 -> 25
1 - 2 + 3 + 4 -> 6
1 - 2 + 3 - 4 -> -2
1 - 2 + 3 / 4 -> -1
1 - 2 + 3 * 4 -> 11
1 - 2 - 3 + 4 -> 0
1 - 2 - 3 - 4 -> -8
1 - 2 - 3 / 4 -> -1
1 - 2 - 3 * 4 -> -13
1 - 2 / 3 + 4 -> 5
1 - 2 / 3 - 4 -> -3
1 - 2 / 3 / 4 -> 1
1 - 2 / 3 * 4 -> 1
1 - 2 * 3 + 4 -> -1
1 - 2 * 3 - 4 -> -9
1 - 2 * 3 / 4 -> 0
1 - 2 * 3 * 4 -> -23
1 / 2 + 3 + 4 -> 7
1 / 2 + 3 - 4 -> -1
1 / 2 + 3 / 4 -> 0
1 / 2 + 3 * 4 -> 12
1 / 2 - 3 + 4 -> 1
1 / 2 - 3 - 4 -> -7
1 / 2 - 3 / 4 -> 0
1 / 2 - 3 * 4 -> -12
1 / 2 / 3 + 4 -> 4
1 / 2 / 3 - 4 -> -4
1 / 2 / 3 / 4 -> 0
1 / 2 / 3 * 4 -> 0
1 / 2 * 3 + 4 -> 4
1 / 2 * 3 - 4 -> -4
1 / 2 * 3 / 4 -> 0
1 / 2 * 3 * 4 -> 0
1 * 2 + 3 + 4 -> 9
1 * 2 + 3 - 4 -> 1
1 * 2 + 3 / 4 -> 2
1 * 2 + 3 * 4 -> 14
1 * 2 - 3 + 4 -> 3
1 * 2 - 3 - 4 -> -5
1 * 2 - 3 / 4 -> 2
1 * 2 - 3 * 4 -> -10
1 * 2 / 3 + 4 -> 4
1 * 2 / 3 - 4 -> -4
1 * 2 / 3 / 4 -> 0
1 * 2 / 3 * 4 -> 0
1 * 2 * 3 + 4 -> 10
1 * 2 * 3 - 4 -> 2
1 * 2 * 3 / 4 -> 1
1 * 2 * 3 * 4 -> 24
计算所有可能的运算符组合，包括用括号分组和重新排序，这种方法更复杂。
答案 2 :(得分：-1)
如果您想将结果打印为字符串，那么您应该这样：
int a=1,b=2,c=3,w=4,result;
char d='+',e='-',f='/',g='*';
//a d b g w =result;
printf("%d %c %d %c %d", a, d, b, g, w);
但是，如果要计算表达式的结果，则需要创建一个函数来执行此操作。类似于this。