这个C++程序为什么不能运行# Programming - 葵花宝典
b*n
1 楼
z*i
2 楼
Could it be:
char *a ---> pointer to read only section
char a[]---> array in stack, initialized with "abcd", which is in read only
section.
【在 b*********n 的大作中提到】
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: a[0] = 'f';
: cout<: return 0;
: }
: compile没问题(Visual Studio 6)
: 如果定义char a[]="abcde";
: 就没问题
char *a ---> pointer to read only section
char a[]---> array in stack, initialized with "abcd", which is in read only
section.
【在 b*********n 的大作中提到】
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: a[0] = 'f';
: cout<: return 0;
: }
: compile没问题(Visual Studio 6)
: 如果定义char a[]="abcde";
: 就没问题
d*a
5 楼
c++? if you really want c++, use string instead of char * and you'll get rid
of most of this kind trouble.
first basic things to use c++ instead of c:
alloc, malloc/free -> new/delete
char [], char* -> string
array -> vector
of most of this kind trouble.
first basic things to use c++ instead of c:
alloc, malloc/free -> new/delete
char [], char* -> string
array -> vector
s9
6 楼
a[0] = 'f'出错是因为它试图去修改一个不允许修改的字符串常量。
char *a="abcde"; a是一个处于堆栈的指针，其值是"abcde"这个字符串常量的地址。
长度是4（32 bit PC）。
char a[]="abcde" a是一个处于堆栈的数组，其长度取决于其所包含的字符串的长度。
【在 b*********n 的大作中提到】
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: a[0] = 'f';
: cout<: return 0;
: }
: compile没问题(Visual Studio 6)
: 如果定义char a[]="abcde";
: 就没问题
char *a="abcde"; a是一个处于堆栈的指针，其值是"abcde"这个字符串常量的地址。
长度是4（32 bit PC）。
char a[]="abcde" a是一个处于堆栈的数组，其长度取决于其所包含的字符串的长度。
【在 b*********n 的大作中提到】
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: a[0] = 'f';
: cout<: return 0;
: }
: compile没问题(Visual Studio 6)
: 如果定义char a[]="abcde";
: 就没问题
c*t
7 楼
check this link
http://c-faq.com/decl/strlitinit.html
【在 b*********n 的大作中提到】
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: a[0] = 'f';
: cout<: return 0;
: }
: compile没问题(Visual Studio 6)
: 如果定义char a[]="abcde";
: 就没问题
http://c-faq.com/decl/strlitinit.html
【在 b*********n 的大作中提到】
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: a[0] = 'f';
: cout<: return 0;
: }
: compile没问题(Visual Studio 6)
: 如果定义char a[]="abcde";
: 就没问题
s*e
8 楼
应该有Warning吧？应该这么写
const char* str="abcd";
const char* str="abcd";
d*y
9 楼
char *a;是一个指针,没有ASSIGN存储空间.
char *a="abcde"; 是给这个指针一个值"abcde","abcde"可能会被TRUNCTED,depending
on your system.
指针和数组是有本质区别的.
char *a="abcde"; 是给这个指针一个值"abcde","abcde"可能会被TRUNCTED,depending
on your system.
指针和数组是有本质区别的.
c*n
10 楼
I just did some test. See the program below:
===============================
#include
#include
#include
int main(int argc, char* argv[])
{
char *a="abcde";
char b[]="abcde";
char* c=(char*)malloc(6);
strcpy(c, a);
printf("%lx, %lx, %lx\n", a, b, c);
printf("%c %c %c\n", a[0], b[0], c[0]);
a[0] = 'f';
printf("%s\n", a);
return 0;
}
==============================
output:
80486d0, bfc74b96, 804b008
a a a
Segmentation fault
char *a="abcde", the compiler generate a str
===============================
#include
#include
#include
int main(int argc, char* argv[])
{
char *a="abcde";
char b[]="abcde";
char* c=(char*)malloc(6);
strcpy(c, a);
printf("%lx, %lx, %lx\n", a, b, c);
printf("%c %c %c\n", a[0], b[0], c[0]);
a[0] = 'f';
printf("%s\n", a);
return 0;
}
==============================
output:
80486d0, bfc74b96, 804b008
a a a
Segmentation fault
char *a="abcde", the compiler generate a str
k*f
11 楼
C老问题拉，去精华区看看吧
sizeof(a)和sizeof(b)是不一样的
【在 c****n 的大作中提到】
: I just did some test. See the program below:
: ===============================
: #include
: #include
: #include
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: char b[]="abcde";
: char* c=(char*)malloc(6);
sizeof(a)和sizeof(b)是不一样的
【在 c****n 的大作中提到】
: I just did some test. See the program below:
: ===============================
: #include
: #include
: #include
: int main(int argc, char* argv[])
: {
: char *a="abcde";
: char b[]="abcde";
: char* c=(char*)malloc(6);
b*y
12 楼
It is a pure C question rather than C++.
char *a = "abcde";
a is not in stack or heap. It is in the memory which is read only. Any
attempt to change the content of a is wrong.
char *a = "abcde";
a is not in stack or heap. It is in the memory which is read only. Any
attempt to change the content of a is wrong.
b*n
13 楼
a is just a pointer. It is initialised to point to read-only memory. Changin
g a is fine, changing the content of memory pointed by a is not okay
【在 b**y 的大作中提到】
: It is a pure C question rather than C++.
: char *a = "abcde";
: a is not in stack or heap. It is in the memory which is read only. Any
: attempt to change the content of a is wrong.
g a is fine, changing the content of memory pointed by a is not okay
【在 b**y 的大作中提到】
: It is a pure C question rather than C++.
: char *a = "abcde";
: a is not in stack or heap. It is in the memory which is read only. Any
: attempt to change the content of a is wrong.
y*a
14 楼
下面是汇编代码，string "abcde" is in readonly section (.rodata). 除非你修改
gcc default linker script, you cannot write to this address
.file "l.c"
.section .rodata
.LC0:
.string "abcde"
.text
.globl main
.type main, @function
main:
pushl %ebp
movl %esp, %ebp
subl $8, %esp
andl $-16, %esp
movl $0, %eax
addl $15, %eax
addl $15, %eax
shrl $4, %eax
sall $4, %eax
gcc default linker script, you cannot write to this address
.file "l.c"
.section .rodata
.LC0:
.string "abcde"
.text
.globl main
.type main, @function
main:
pushl %ebp
movl %esp, %ebp
subl $8, %esp
andl $-16, %esp
movl $0, %eax
addl $15, %eax
addl $15, %eax
shrl $4, %eax
sall $4, %eax
d*a
15 楼
niu!
【在 y*****a 的大作中提到】
: 下面是汇编代码，string "abcde" is in readonly section (.rodata). 除非你修改
: gcc default linker script, you cannot write to this address
: .file "l.c"
: .section .rodata
: .LC0:
: .string "abcde"
: .text
: .globl main
: .type main, @function
: main:
【在 y*****a 的大作中提到】
: 下面是汇编代码，string "abcde" is in readonly section (.rodata). 除非你修改
: gcc default linker script, you cannot write to this address
: .file "l.c"
: .section .rodata
: .LC0:
: .string "abcde"
: .text
: .globl main
: .type main, @function
: main:
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