如何将struct中的数组更改为C中的指针
我的大脑和功能:如何将struct中的数组更改为C中的指针,c,C,我的大脑和功能: #define MAX 4096 typedef struct{ char comment[40]; int nbpts; float time[4096]; float value[4096]; } trace; void simuTrace(int tmax, float dt, float params[], trace* uneTrace){ int i = 0; float v = 0, w = 0, dv = 0, dw =
#define MAX 4096
typedef struct{
char comment[40];
int nbpts;
float time[4096];
float value[4096];
} trace;
void simuTrace(int tmax, float dt, float params[], trace* uneTrace){
int i = 0;
float v = 0, w = 0, dv = 0, dw = 0, t = 0;
while (t<tmax && i < MAX){
dv = (params[0]-v)*(v-1)*v - w;
dw = params[4]*(params[1]*v-params[2]*w-params[3]);
v += dv*dt;
w += dw*dt;
uneTrace->time[i] = t;
uneTrace->value[i] = v;
i++;
t += dt;
}
uneTrace->nbpts = i+1;
uneTrace->comment[40]= "CommentaireDeLaTrace";
}
使用
malloc
typedef struct{
char comment[40];
int nbpts;
float *time;
float *value;
} trace;
void simuTrace(int tmax, float dt, float params[], trace* uneTrace){
int i = 0;
float v = 0, w = 0, dv = 0, dw = 0, t = 0;
int count = tmax/dt;
uneTrace->time = malloc(count * sizeof(float));
uneTrace->value = malloc(count * sizeof(float));
while (i < count){
dv = (params[0]-v)*(v-1)*v - w;
dw = params[4]*(params[1]*v-params[2]*w-params[3]);
v += dv*dt;
w += dw*dt;
uneTrace->time[i] = t;
uneTrace->value[i] = v;
i++;
t += dt;
}
uneTrace->nbpts = i+1;
strcpy(uneTrace->comment, "CommentaireDeLaTrace");
}
typedef结构{
char评论[40];
int nbpts;
浮动*时间;
浮动*值;
}痕迹;
void simuTrace(int-tmax,float-dt,float-params[],trace*uneTrace){
int i=0;
浮点数v=0,w=0,dv=0,dw=0,t=0;
整数计数=tmax/dt;
uneTrace->time=malloc(count*sizeof(float));
uneTrace->value=malloc(count*sizeof(float));
而(我<计数){
dv=(参数[0]-v)*(v-1)*v-w;
dw=params[4]*(params[1]*v-params[2]*w-params[3]);
v+=dv*dt;
w+=dw*dt;
uneTrace->time[i]=t;
uneTrace->value[i]=v;
i++;
t+=dt;
}
uneTrace->nbpts=i+1;
strcpy(uneTrace->comment,“commentairederatrace”);
}
您没有收到编译器警告吗?那么uneTrace->comment[40]=“commentairedelaterace”呢
这会将不兼容的值(char*
指针指向char
)写入数组边界之外。使用strcpy
复制字符串。您不能分配给数组,数组应该是strcpy(uneTrace->comment,“commentairedelaterace”)
请使用中的MAX
来定义MAX 4096
而不是硬编码4096
。为什么设置变量count=tmax/dt?请解释一下,您的循环将t
从0
增加到tmax
,增加dt
。这意味着它执行tmax/dt
迭代,这就是它将放入数组的元素数。简单的算术。
typedef struct{
char comment[40];
int nbpts;
float *time;
float *value;
} trace;
void simuTrace(int tmax, float dt, float params[], trace* uneTrace){
int i = 0;
float v = 0, w = 0, dv = 0, dw = 0, t = 0;
int count = tmax/dt;
uneTrace->time = malloc(count * sizeof(float));
uneTrace->value = malloc(count * sizeof(float));
while (i < count){
dv = (params[0]-v)*(v-1)*v - w;
dw = params[4]*(params[1]*v-params[2]*w-params[3]);
v += dv*dt;
w += dw*dt;
uneTrace->time[i] = t;
uneTrace->value[i] = v;
i++;
t += dt;
}
uneTrace->nbpts = i+1;
strcpy(uneTrace->comment, "CommentaireDeLaTrace");
}