分段错误（核心转储）的C运行时错误 [英] Segmentation fault (core dumped) runtime error in C
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问题描述
所以我新的C，只是写了我的第一个程序的语言和不断收到分段错误，当我尝试运行它。我敢肯定，有我在整个code由多个小失误。我已经通过它去，我不能揣摩出我的错误是。这里是我的code：
// $编号：crpn.c，1.1版2013年10月22日13：28：04-07 - - $＃包括LT＆;＆ASSERT.H GT;
＃包括LT＆;＆libgen.h GT;
＃包括LT＆;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;INT EXIT_STATUS = EXIT_SUCCESS;
的#define EMPTY（-1）
＃定义尺寸16typedef结构栈堆栈;
结构栈{
INT榜首;
INT能力;
INT大小;
双号[SIZE]
};无效bad_operator（为const char * OPER）{
fflush（NULL）;
fprintf中（标准错误，OPER = \\％s \\的\\ n，OPER）;
fflush（NULL）;
EXIT_STATUS = EXIT_FAILURE;
的printf（％S：invaild运营商\\ n，OPER）;
}无效的push（堆栈* the_stack，双号）{
如果（the_stack-＆GT;大小== the_stack-＆GT;能力）{
的printf（％A：堆栈溢出号）;
}
其他{
the_stack-＆GT;数字[the_stack-＆GT;大小++] =号;
}
}无效do_binop（堆栈* the_stack，焦炭OPER）{
如果（（the_stack-＆GT;顶部）。1）{
的printf（OPER = \\％C \\：堆栈下溢\\ n，OPER）;
}
其他{
双右= the_stack-＆GT;数字[the_stack-＆GT; size--]。
双击左键= the_stack-＆GT;数字[the_stack-＆GT; size--]。
开关（OPER）{
案例'+'：推（the_stack，+左右）;打破;
案例' - '：推（the_stack，左 - 右）;打破;
案例'*'：推（the_stack，左*右）;打破;
案例'/'：推（the_stack，左/右）;打破;
}
}
}无效do_print（堆栈* the_stack）{
如果（the_stack-＆GT;顶== -1）{
的printf（堆栈是空\\ n）;
}
其他{
INT POS;
对于（POS = 0; POS＆LT; = the_stack-＆GT;顶++ POS）{
的printf（％A \\ N，the_stack-＆GT;数字[POS]）;
}
}
}无效do_clear（堆栈* the_stack）{
the_stack-＆GT;顶= -1;
}无效do_operator（堆栈* the_stack，为const char * OPER）{
开关（OPER [0]）{
案例'+'：do_binop（the_stack，'+'）;打破;
案例' - '：do_binop（the_stack，' - '）;打破;
情况下*：do_binop（the_stack，'*'）;打破;
案例'/'：do_binop（the_stack，'/'）;打破;
案;：do_print（the_stack）;打破;
案例'@'：do_clear（the_stack）;打破;
默认：bad_operator（OPER）;打破;
}
}INT主（INT ARGC，字符** argv的）{
如果（argc个！= 1）{
fprintf中（标准错误，用法：％S \\ n，基名（的argv [0]））;
fflush（NULL）;
出口（EXIT_FAILURE）;
}
堆叠the_stack;
the_stack.top =空的;
字符缓冲区[1024];
为（;;）{
INT scanrc = scanf函数（％1023s，缓冲区）;
如果（scanrc == EOF）破;
断言（scanrc == 1）;
如果（缓冲[0] =='＃'）{
scanrc = scanf函数（％1023 [^ \\ n]，缓冲区）;
继续;
}
字符* endptr;
双数=的strtod（缓冲，＆安培; endptr）;
如果（* endptr =='\\ 0'）{
推（安培; the_stack，号码）;
}否则如果（缓冲[1]！='\\ 0'）{
bad_operator（缓冲液）;
}其他{
do_operator（安培; the_stack，缓冲区）;
}
}
返回EXIT_STATUS;
}
解决方案
让我教你以渔：
一个调试器会告诉你到底何处有故障。如果您使用的是IDE（X code，Eclipse中，VS），它有一个很好的接口之一，你应该使用。如果不是：
与 -g 开关编译程序： gcc的-g我的code.C 。这将调试信息添加到可执行文件（使调试器给你更好的信息）。
$ GDB my_executable
...
＆GT;跑
...
分段故障
＆GT;哪里
这会给你的准确位置（这在其行号的功能）。
Hi so I am new to C and just wrote my first program in the language and keep getting the segmentation fault error when I try to run it. I'm sure there are multiple small mistakes I have made throughout the code. I have gone through it and I can't figure out where my mistake is. Here is my code:
// $Id: crpn.c,v 1.1 2013-10-22 13:28:04-07 - - $
#include <assert.h>
#include <libgen.h>
#include <stdio.h>
#include <stdlib.h>
int exit_status = EXIT_SUCCESS;
#define EMPTY (-1)
#define SIZE 16
typedef struct stack stack;
struct stack {
int top;
int capacity;
int size;
double numbers[SIZE];
};
void bad_operator (const char *oper) {
fflush (NULL);
fprintf (stderr, "oper=\"%s\"\n", oper);
fflush (NULL);
exit_status = EXIT_FAILURE;
printf("%s: invaild operator\n", oper);
}
void push (stack *the_stack, double number) {
if (the_stack->size == the_stack->capacity) {
printf("%a:stack overflow", number);
}
else {
the_stack->numbers[the_stack->size++]=number;
}
}
void do_binop (stack *the_stack, char oper) {
if ((the_stack->top)<1) {
printf("oper=\"%c\":stack underflow\n", oper);
}
else {
double right = the_stack->numbers[the_stack->size--];
double left = the_stack->numbers[the_stack->size--];
switch (oper) {
case '+': push (the_stack, left + right); break;
case '-': push (the_stack, left - right); break;
case '*': push (the_stack, left * right); break;
case '/': push (the_stack, left / right); break;
}
}
}
void do_print (stack *the_stack) {
if (the_stack->top == -1) {
printf("stack is empty\n");
}
else {
int pos;
for (pos = 0; pos <= the_stack->top; ++pos) {
printf("%a\n",the_stack->numbers[pos]);
}
}
}
void do_clear (stack *the_stack) {
the_stack->top = -1;
}
void do_operator (stack *the_stack, const char *oper) {
switch (oper[0] ) {
case '+': do_binop (the_stack, '+'); break;
case '-': do_binop (the_stack, '-'); break;
case '*': do_binop (the_stack, '*'); break;
case '/': do_binop (the_stack, '/'); break;
case ';': do_print (the_stack); break;
case '@': do_clear (the_stack); break;
default : bad_operator (oper); break;
}
}
int main (int argc, char **argv) {
if (argc != 1) {
fprintf (stderr, "Usage: %s\n", basename (argv[0]));
fflush (NULL);
exit (EXIT_FAILURE);
}
stack the_stack;
the_stack.top = EMPTY;
char buffer[1024];
for (;;) {
int scanrc = scanf ("%1023s", buffer);
if (scanrc == EOF) break;
assert (scanrc == 1);
if (buffer[0] == '#') {
scanrc = scanf ("%1023[^\n]", buffer);
continue;
}
char *endptr;
double number = strtod (buffer, &endptr);
if (*endptr == '\0') {
push (&the_stack, number);
}else if (buffer[1] != '\0') {
bad_operator (buffer);
}else {
do_operator (&the_stack, buffer);
}
}
return exit_status;
}
解决方案
Let me "teach you to fish":
A debugger will tell you exactly where the fault is. If you're using an IDE (Xcode, Eclipse, VS) it has a nice interface to one and you should use that. If not:
Compile your program with the -g switch: gcc -g mycode.c. This adds debugging information to the executable (makes the debugger give you much better info).
$ gdb my_executable
...
> run
...
Segmentation fault
> where
This will give you the exact location (which function on which line number).
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