分配中不兼容的类型 [英] incompatible types in assignment
问题描述
行号中的代码有什么问题。 7?!
#cat test3.c
#include< stdio.h>
int main (int argc,char * argv [])
{
int ia1 [10] = {0,1,2};
int ia2 [10];
ia2 = ia1;
printf("%d \ nn",ia2 [2]);
返回0;
}
#gcc test3.c
test3.c:在函数`main'中':
test3.c:7:作业中不兼容的类型
#
Whats wrong with the code in line no. 7?!
#cat test3.c
#include <stdio.h>
int main(int argc, char *argv[])
{
int ia1[10] = {0, 1, 2};
int ia2[10];
ia2 = ia1;
printf("%d\n", ia2[2]);
return 0;
}
#gcc test3.c
test3.c: In function `main'':
test3.c:7: incompatible types in assignment
#
推荐答案
v4vijayakumar说:
v4vijayakumar said:
行代码中的代码有什么问题。 7?!
#cat test3.c
#include< stdio.h>
int main(int argc,char * argv [])
{
int ia1 [10] = {0,1,2};
int ia2 [10];
ia2 = ia1;
printf(" ;%d \ n",ia2 [2]);
返回0;
}
#gcc test3.c
test3.c:在函数中main'':
test3.c:7:赋值中不兼容的类型
Whats wrong with the code in line no. 7?!
#cat test3.c
#include <stdio.h>
int main(int argc, char *argv[])
{
int ia1[10] = {0, 1, 2};
int ia2[10];
ia2 = ia1;
printf("%d\n", ia2[2]);
return 0;
}
#gcc test3.c
test3.c: In function `main'':
test3.c:7: incompatible types in assignment
你不能用C语言分配数组。它们是左值,但是不是可修改的
lvalues。
修复:
#include< string.h> <然后,如果两个阵列具有相同的大小和类型,你可以用这个来替换你的作业尝试。
memcpy(ia2,ia1,sizeof ia2);
-
Richard Heathfield
" Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件:rjh在上面的域名(但显然放弃了www)
You can''t assign to arrays in C. They are lvalues, but not "modifiable
lvalues".
Fix:
#include <string.h>
and then, provided the two arrays are of the same size and type, you can
replace your assignment attempt with this::
memcpy(ia2, ia1, sizeof ia2);
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
另外。
>
int main(int argc,char * argv [])
{
int ia1 [10] = {0,1,2};
int * ia2;
ia2 = ia1;
printf("%d \ n",* ia2 + 2) ;
返回0;
}
v4vijayakumar写道:
Also.
int main(int argc, char *argv[])
{
int ia1[10] = {0, 1, 2};
int *ia2;
ia2 = ia1;
printf("%d\n",*ia2+2);
return 0;
}
v4vijayakumar wrote:
排队的代码有什么问题。 7?!
#cat test3.c
#include< stdio.h>
int main(int argc,char * argv [])
{
int ia1 [10] = {0,1,2};
int ia2 [10];
ia2 = ia1;
printf(" ;%d \ n",ia2 [2]);
返回0;
}
#gcc test3.c
test3.c:在函数中main'':
test3.c:7:作业中不兼容的类型
#
Whats wrong with the code in line no. 7?!
#cat test3.c
#include <stdio.h>
int main(int argc, char *argv[])
{
int ia1[10] = {0, 1, 2};
int ia2[10];
ia2 = ia1;
printf("%d\n", ia2[2]);
return 0;
}
#gcc test3.c
test3.c: In function `main'':
test3.c:7: incompatible types in assignment
#
#include< stdio.h>
int main(int argc,char * argv [])
{
int ia1 [10] = {0,1,2};
int * ia2;
ia2 = ia1;
printf("% d \ n,*(ia2 + 2));
返回0;
}
v4vijayakumar写道:
#include <stdio.h>
int main(int argc, char *argv[])
{
int ia1[10] = {0, 1, 2};
int *ia2;
ia2 = ia1;
printf("%d\n",*(ia2+2));
return 0;
}
v4vijayakumar wrote:
行中没有错误的代码。 7?!
#cat test3.c
#include< stdio.h>
int main(int argc,char * argv [])
{
int ia1 [10] = {0,1,2};
int ia2 [10];
ia2 = ia1;
printf(" ;%d \ n",ia2 [2]);
返回0;
}
#gcc test3.c
test3.c:在函数中main'':
test3.c:7:作业中不兼容的类型
#
Whats wrong with the code in line no. 7?!
#cat test3.c
#include <stdio.h>
int main(int argc, char *argv[])
{
int ia1[10] = {0, 1, 2};
int ia2[10];
ia2 = ia1;
printf("%d\n", ia2[2]);
return 0;
}
#gcc test3.c
test3.c: In function `main'':
test3.c:7: incompatible types in assignment
#
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